We have a grid with N rows and N columns. The cell at the i-th row and j-th column is denoted (i, j).

Initially, M of the cells are painted black, and all other cells are white. Specifically, the cells (a_1, b_1), (a_2, b_2), ..., (a_M, b_M) are painted black.

Snuke will try to paint as many white cells black as possible, according to the following rule:

* If two cells (x, y) and (y, z) are both black and a cell (z, x) is white for integers 1≤x,y,z≤N, paint the cell (z, x) black.



Find the number of black cells when no more white cells can be painted black.

Constraints

* 1≤N≤10^5
* 1≤M≤10^5
* 1≤a_i,b_i≤N
* All pairs (a_i, b_i) are distinct.

Input

The 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 the number of black cells when no more white cells can be painted black.

Examples

Input

3 2
1 2
2 3


Output

3


Input

2 2
1 1
1 2


Output

4


Input

4 3
1 2
1 3
4 4


Output

3
#include<bits/stdc++.h>
#include<vector>
#define N 100005
#define LL long long
using namespace std;
vector<int> G[N],E[N];
int n,m,f[N],col[N],cnt[3],sz[N],xzy,tot; LL ans;
int find(int x){ return f[x] ? f[x]=find(f[x]) : x;}
void dfs(int t)
{
	int i; cnt[col[t]]++,tot+=G[t].size();
	for(i=0;i<G[t].size();i++){
		if(col[G[t][i]]!=-1){
			if(col[G[t][i]]!=((col[t]+E[t][i]+3)%3)) xzy=0;
			continue;
		  }
		col[G[t][i]]=(col[t]+E[t][i]+3)%3;
		dfs(G[t][i]);
	  }
}
void solve(int t)
{
	col[t]=cnt[0]=cnt[1]=cnt[2]=tot=0,xzy=1;
	dfs(t);
	if(!xzy) ans+=1LL*sz[t]*sz[t];
	else{
		if(cnt[0]&&cnt[1]&&cnt[2])
			ans+=1LL*cnt[0]*cnt[1]+1LL*cnt[1]*cnt[2]+1LL*cnt[2]*cnt[0];
		else ans+=tot/2;
	  }
}
int main()
{
	int i,x,y;
	scanf("%d %d",&n,&m);
	memset(col,-1,sizeof(col));
	for(i=1;i<=m;i++){
		scanf("%d %d",&x,&y);
		G[x].push_back(y);
		E[x].push_back(+1);
		G[y].push_back(x);
		E[y].push_back(-1);
		x=find(x),y=find(y);
		if(x!=y) f[x]=y;
	  }
	for(i=1;i<=n;i++) sz[find(i)]++;
	for(i=1;i<=n;i++)
		if(find(i)==i) solve(i);
	cout<<ans;
	return 0;
}
