//2538224 2010-06-13 15:25:40 Accepted 1811 31MS 640K 2121 B C++ T&T
//把=號用並查集處理以後(所有同一集合的數字用根數字來表示)就是顯而易見的拓撲排序了
#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#define MAX 10010
using namespace std;
vector<int>cost[MAX];
int deep[MAX],count[MAX],save[20005][2];
int find(int x)
{
int r,j;
r = x;
while(r != p[r])
{
j = r;
r = p[r];
p[j] = p[r]; //壓縮路徑
}
return r;
}
void merge(int x,int y)
{
int fx,fy;
fx = find(x);
fy = find(y);
if(fx != fy)
{
if(deep[fx] > deep[fy])
{
p[fy] = fx;
} //深度小的數加到深度大的樹下可以儘量減少樹的深度減少查詢時間
else
{
p[fx] = fy;
if(deep[fx] == deep[fy])
{
deep[fy]++; //深度相同的樹相加 數的深度纔會增加
}
}
}
}
int Toposort(int n)
{
int top = -1,i,k,num = 0,j,x;
int res = 1;
for(i = 0; i < n; i++)
{
if(count[i] == 0 && find(i) == i)
{
count[i] = top;
top = i;
num++;
}
}
if(num > 1)
{
res = 2;
}//uncertain
else if(num == 0)
{
return 3;
}
for(i = 0; i < n; i++)
{
num = 0;
if(find(i) == i) //這個地方糾結了一個夜 忘了考慮本題不是所有的元素都拿來排序的直接抄模板錯了6 7次
{//數組下標模擬堆棧
if(top == -1)//循環次數少於圖的點數 即當所有的點入度都不爲0卻還有點未取出 就是有環存在的情況
{
//res = 3;
return 3; //衝突優於信息不全
}
else
{
j = top;
top = count[top];
for(k = 0; k < cost[j].size(); k++)
{
x = cost[j][k];
if((--count[x]) == 0)
{
count[x] = top;
top = x;
num++;
}
}
if(num > 1) //uncertain
{
res = 2;
}
}
}
}
return res;
}
int main()
{
int n,m,i,a,b,t,d,ans;
char sign;
while(scanf("%d%d",&n,&m) != EOF)
{
d = 0;
// memset(count,0,sizeof(count));
for(i = 0; i <= n; i++)
{
count[i] = 0;
deep[i] = 1;
p[i] = i;
cost[i].clear();
}
for(i = 0; i < m; i++)
{
scanf("%d%s%d",&a,&sign,&b);
if(sign != '=')
{
if(sign == '>')
{
//count[save[i][0]]++;
t = a;
a = b;
b = t;
} //這裏先交換保障 save[i][1]要大以便後面的操作
save[d][0] = a;save[d++][1] = b;
}
else
{
merge(a,b); //等於用並查集處理
}
}
for(i = 0; i < d; i++)
{
a = find(save[i][0]);
b = find(save[i][1]);
cost[a].push_back(b); //建邊
count[b]++;
}
ans = 1;
ans = Toposort(n);
if(ans == 1)
{
printf("OK/n");
}
else if(ans == 2)
{
printf("UNCERTAIN/n");
}
else if(ans == 3)
{
printf("CONFLICT/n");
}
}
}