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<algorithm>
#include<cstdio>
#define M 200000
#define N 1001
using namespace std;
int a[N],b[M],c[M],d[M],e[N][N],f[N][N],m,n;bool x[N];
inline bool cmp(int u,int v){return b[u]==b[v]?c[u]<c[v]:b[u]<b[v];}
void dfs(int u,int*w){for(int i=a[u-1];i<a[u];i++)if(!w[b[i]])w[b[i]]=w[u],dfs(b[i],w);}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++)scanf("%d%d",b+i,c+i),a[b[d[i]=i]]++;
	sort(d,d+m,cmp);
	for(int i=1;i<=n;i++)a[i]+=a[i-1],e[i][i]=f[i][i]=-1;
	for(int i=0;i<m;i++)b[i]=c[d[i]];
	for(int i=0;i<m;i++)c[i]=d[i];
	for(int i=1;i<=n;i++)for(int j=a[i-1];j<a[i];j++)if(!e[i][b[j]])dfs(e[i][b[j]]=b[j],e[i]);
	for(int i=1;i<=n;i++)for(int j=a[i];a[i-1]<j--;)if(!f[i][b[j]])dfs(f[i][b[j]]=b[j],f[i]);
	for(int i=1;i<=n;i++)for(int j=a[i-1];j<a[i];j++)x[c[j]]=(bool)e[b[j]][i]^e[i][b[j]]==f[i][b[j]];
	for(int i=0;i<m;i++)puts(x[i]?"same":"diff");
	return 0;
}