Problem:
Score : 1100 points

Problem Statement

You are given a directed graph with N vertices and M edges.
The vertices are numbered 1, 2, ..., N, and the edges are numbered 1, 2, ..., M.
Edge i points from Vertex a_i to Vertex b_i.

For each edge, determine whether the reversion of that edge would change the number of the strongly connected components in the graph.

Here, the reversion of Edge i means deleting Edge i and then adding a new edge that points from Vertex b_i to Vertex a_i.

Constraints

2 \leq N \leq 1000

1 \leq M \leq 200,000

1 \leq a_i, b_i \leq N

a_i \neq b_i

If i \neq j, then a_i \neq a_j or b_i \neq b_j.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print M lines. In the i-th line, if the reversion of Edge i would change the number of the strongly connected components in the graph, print diff; if it would not, print same.

Sample Input 1

3 3
1 2
1 3
2 3

Sample Output 1

same
diff
same

The number of the strongly connected components is 3 without reversion of edges, but it will become 1 if Edge 2 is reversed.

Sample Input 2

2 2
1 2
2 1

Sample Output 2

diff
diff

Reversion of an edge may result in multiple edges in the graph.

Sample Input 3

5 9
3 2
3 1
4 1
4 2
3 5
5 3
3 4
1 2
2 5

Sample Output 3

same
same
same
same
same
diff
diff
diff
diff

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
vector<int> e[1005];
int cl[1005][1005][2],u[200005],v[200005];
void dfs(int x,int ro,int c,int t){
	if(cl[ro][x][t])return;
	cl[ro][x][t]=c;
	for(int i=0;i<e[x].size();i++)if(!cl[ro][e[x][i]][t])dfs(e[x][i],ro,c,t);
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u[i],&v[i]);
		e[u[i]].push_back(v[i]);
	}
	for(int i=1;i<=n;i++){
		cl[i][i][0]=i;
		for(int j=0;j<e[i].size();j++){
			dfs(e[i][j],i,e[i][j],0);
		}
		cl[i][i][1]=i;
		for(int j=(int)e[i].size()-1;j>=0;j--){
			dfs(e[i][j],i,e[i][j],1);
		}
	}
	for(int i=1;i<=m;i++){
		if(cl[v[i]][u[i]][0]){
			if(cl[u[i]][v[i]][0]!=v[i]||cl[u[i]][v[i]][1]!=v[i])puts("same");
			else puts("diff");
		}else{
			if(cl[u[i]][v[i]][0]!=v[i]||cl[u[i]][v[i]][1]!=v[i])puts("diff");
			else puts("same");
		}
	}
	return 0;
}