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 <bits/stdc++.h>
using namespace std;
const int Maxn=1005,Maxm=200005;
int vis[Maxn][Maxn],lst1[Maxn][Maxn],lst2[Maxn][Maxn],ans[Maxm];
vector<pair<int,int>>e[Maxn];
void dfs1(int x,int s){
    vis[s][x]=true;
    for(auto i:e[x]){
        int y=i.first;
        if(vis[s][y])continue;
        dfs1(y,s);
    }
}
void dfs2(int x,int s,int lst[][Maxn]){
    for(auto i:e[x]){
        int y,id;tie(y,id)=i;
        if(!lst[s][y]&&y!=s){
            lst[s][y]=id;
            dfs2(y,s,lst);
        }
    }
}
int main(){
    int n,m;scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int x,y;scanf("%d%d",&x,&y);
        e[x].emplace_back(y,i);
    }
    for(int i=1;i<=n;i++)
        dfs1(i,i);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)reverse(e[j].begin(),e[j].end());
        dfs2(i,i,lst1);
        for(int j=1;j<=n;j++)reverse(e[j].begin(),e[j].end());
        dfs2(i,i,lst2);
        for(auto o:e[i]){
            int y,id;tie(y,id)=o;
            ans[id]=vis[y][i]^(lst1[i][y]!=id||lst2[i][y]!=id);
        }
    }
    for(int i=1;i<=m;i++)
        puts(ans[i]?"diff":"same");
    return 0;
}