Berland Beauty

There are n railway stations in Berland. They are connected to each other by n−1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.

You have a map of that network, so for each railway section you know which stations it connects.

Each of the n−1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don’t know them. All these values are from 1 to $10^6$ inclusive.

You asked m passengers some questions: the j-th one told you three values:

his departure station $a_j$;
his arrival station $b_j$;
minimum scenery beauty along the path from $a_j$ to $b_j$ (the train is moving along the shortest path from $a_j$ to $b_j$).
You are planning to update the map and set some value $f_i$ on each railway section — the scenery beauty. The passengers’ answers should be consistent with these values.

Print any valid set of values $f_1,f_2,…,f_{n−1}$, which the passengers’ answer is consistent with or report that it doesn’t exist.

Input
The first line contains a single integer $n (2≤n≤5000)$ — the number of railway stations in Berland.

The next $n−1$ lines contain descriptions of the railway sections: the i-th section description is two integers $x_i$ and $y_i$ $(1≤x_i,y_i≤n,x_i≠y_i)$, where $x_i$ and $y_i$ are the indices of the stations which are connected by the i-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.

The next line contains a single integer $m (1≤m≤5000)$ — the number of passengers which were asked questions. Then m lines follow, the j-th line contains three integers $a_j$, $b_j$ and $g_j$ $(1≤a_j,b_j≤n; a_j≠b_j; 1≤g_j≤10^6)$ — the departure station, the arrival station and the minimum scenery beauty along his path.

Output
If there is no answer then print a single integer $-1$.

Otherwise, print $n−1$ integers $f_1,f_2,…,f_{n−1} (1≤f_i≤10^6)$, where $f_i$ is some valid scenery beauty along the i-th railway section.

If there are multiple answers, you can print any of them.

Examples

input
4
1 2
3 2
3 4
2
1 2 5
1 3 3
output
5 3 5
input
6
1 2
1 6
3 1
1 5
4 1
4
6 1 3
3 4 1
6 5 2
1 2 5
output
5 3 1 2 1 
input
6
1 2
1 6
3 1
1 5
4 1
4
6 1 1
3 4 3
6 5 3
1 2 4
output
-1

將詢問按g從小到大排序，然後對於每個詢問，將對應路徑中的邊改爲g，最後再檢查一下。

#include<bits/stdc++.h>

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int,int>
#define mii unordered_map<int,int>
#define msi unordered_map<string,int>
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<<x<<endl
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 5e3 + 10;
int head[N], ver[N << 1], Next[N << 1], edge[N << 1], tot = 1;
int n, m, fa[N], ans[N];
struct Q {
    int a, b, g;
};

bool cmp(Q a, Q b) {
    return a.g < b.g;
}

vector<Q> seq;
namespace LCA {
    int t, f[N][20], d[N];

    void dfs(int x) {
        for (int i = head[x]; i; i = Next[i]) {
            int y = ver[i];
            if (d[y])continue;
            d[y] = d[x] + 1;
            f[y][0] = x;
            fa[y] = i ^ 1;
            for (int j = 1; j <= t; j++)
                f[y][j] = f[f[y][j - 1]][j - 1];
            dfs(y);
        }
    }

    inline int lca(int x, int y) {
        if (d[x] > d[y])swap(x, y);
        for (int i = t; i >= 0; i--)
            if (d[f[y][i]] >= d[x])
                y = f[y][i];
        if (x == y)
            return x;
        for (int i = t; i >= 0; i--)
            if (f[x][i] != f[y][i])
                x = f[x][i], y = f[y][i];
        return f[x][0];
    }

};

inline void add(int x, int y, int z) {
    ver[++tot] = y;
    edge[tot] = z;
    Next[tot] = head[x];
    head[x] = tot;
}

inline void read() {
    n = qr();
    repi(i, 1, n - 1) {
        int x = qr(), y = qr();
        add(x, y, i), add(y, x, i);
    }
    m = qr();
}

void out() {
    repi(i, 1, n - 1)printf("%d%c", ans[i], ce(i, n - 1));
}

int main() {
    read();
    LCA::d[1] = 1;
    LCA::t = log2(n) + 1;
    LCA::dfs(1);
    repi(i, 1, n - 1)ans[i] = 1;
    repi(i, 1, m) {
        int a = qr(), b = qr(), g = qr();
        seq.pb({a, b, g});
    }
    sort(seq.begin(), seq.end(), cmp);
    for (auto it:seq) {
        int a = it.a, b = it.b, g = it.g;
        int lca = LCA::lca(a, b);
        while (a != lca)ans[edge[fa[a]]] = g, a = ver[fa[a]];
        while (b != lca)ans[edge[fa[b]]] = g, b = ver[fa[b]];
    }
    for (auto it:seq) {
        int a = it.a, b = it.b, g = it.g;
        int lca = LCA::lca(a, b), mn = INF;
        while (a != lca)mn = min(mn, ans[edge[fa[a]]]), a = ver[fa[a]];
        while (b != lca)mn = min(mn, ans[edge[fa[b]]]), b = ver[fa[b]];
        if (mn != g) {
            puts("-1");
            return 0;
        }
    }
    repi(i, 1, n - 1)printf("%d%c", ans[i], ce(i, n - 1));
    return 0;
}