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;

#define fi first
#define se second
#define pb push_back
#define mp make_pair

typedef pair<int, int> ii;

const int N = 1005;

vector<int> graph[N];

int n, m, l[200005], r[200005], cnt[N][N], vis[N][N];

void dfs(int u, int p, int root, int mark) {
    if(u == root) return ;
    if(cnt[root][u] >= 2) return ;
    if(vis[root][u] == mark) return ;
    vis[root][u] = mark; cnt[root][u]++;
    for(int i = 0; i < graph[u].size(); i++) {
        int v = graph[u][i];
        if(v == p) continue;
        dfs(v, u, root, mark);
    }
}

int main() {
    cin.tie(0), ios::sync_with_stdio(0);
    cin >> n >> m;
    for(int i = 1; i <= m; i++) {
        cin >> l[i] >> r[i];
        graph[l[i]].pb(r[i]);
    }
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j < graph[i].size(); j++) {
            int v = graph[i][j];
            dfs(v, i, i, v);
        }
    }
    for(int i = 1; i <= m; i++) {
        if(cnt[l[i]][r[i]] >= 2) {
            if(cnt[r[i]][l[i]] == 0)
                cout << "diff\n";
            else
                cout << "same\n";
        }
        else {//1
            if(cnt[r[i]][l[i]] >= 1)
                cout << "diff\n";
            else
                cout << "same\n";
        }
    }
}