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: Runtime Error)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N=1005;

int n,m;
int ans[N],t[N][N],tt[N],b[N*200],a[N*200];

int D,del,clk,vis[N];
void dfs(int u)
{
	if (vis[u]) return;
	if (u==del) return;
	vis[u]=clk;
	for (int i=1;i<=tt[u];i++) dfs(b[t[u][i]]);
}

void work(int u)
{
	del=u;
	memset(vis,0,sizeof(vis));
	for (int v,i=1;i<=tt[u];i++)
	{
		clk=v=b[t[u][i]];
		dfs(v);
	}
	for (int v,i=1;i<=tt[u];i++) 
	{
		v=b[t[u][i]];
		if (v!=vis[v]) ans[t[u][i]]=1;
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int u,v,i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		a[i]=u,b[i]=v;
		t[u][++tt[u]]=i;
	}
	
	for (int u=1;u<=n;u++)
	{
		work(u);
		for (int i=1;i<=tt[u]/2;i++) swap(t[u][i],t[u][ tt[u]-i+1 ]);
		work(u);
	}

	for (int u=1;u<=n;u++)
	{
		memset(vis,0,sizeof(vis));
		clk=1;
		del=0;
		dfs(u);
		for (int i=1;i<=m;i++) if (b[i]==u) ans[i]^=vis[a[i]];
	}

	for (int i=1;i<=m;i++) if (ans[i]) puts("diff");else puts("same");
	return 0;
}