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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int n,m,ind,cnt,top;
int vis[N],que[N],stk[N],bl[N],dfn[N],low[N],u[N*200],v[N*200];
bool ok[N*200],ins[N];
vector<int> ve[N],id[N];
inline void tarjan(int x){
	stk[++top]=x; ins[x]=1; dfn[x]=low[x]=++ind;
	for(auto to:ve[x])
		if(!dfn[to]) tarjan(to),low[x]=min(low[x],low[to]);
		else if(ins[to]) low[x]=min(low[x],dfn[to]);
	if(low[x]==dfn[x]){
		int y; ++cnt;
		do y=stk[top--],ins[y]=0,bl[y]=cnt;
		while(y!=x);
	}
}
inline void solve(int rt){
	int head=1,tail=0; vis[rt]=++ind;
	for(int i=0;i<ve[rt].size();++i){
		ok[id[rt][i]]|=vis[ve[rt][i]]==ind;
		if(vis[ve[rt][i]]!=ind) vis[ve[rt][i]]=ind,que[++tail]=ve[rt][i];
		while(head<=tail){
			int x=que[head++];
			for(auto to:ve[x]) if(vis[to]!=ind) vis[to]=ind,que[++tail]=to;
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,a,b;i<=m;++i) scanf("%d%d",&a,&b),ve[a].push_back(b),id[a].push_back(i),u[i]=a,v[i]=b;
	for(int i=1;i<=n;++i) if(!dfn[i]) tarjan(i);
	for(int i=1;i<=n;++i) solve(i),reverse(ve[i].begin(),ve[i].end()),reverse(id[i].begin(),id[i].end()),solve(i);
	for(int i=1;i<=m;++i){
		if(bl[u[i]]==bl[v[i]]) puts(ok[i]?"same":"diff");
		else puts(ok[i]?"diff":"same");
	}
	return 0;
}