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>
#define N 100100
using namespace std;
int f[N];
int in[N],out[N];
int a[N],b[N];
bool vis[N];
int c[N];
vector<int> g[N],r[N];
void DFS(int u, bool &sl, int sz[], int col){
	vis[u]=true;
	sz[col]++;
	c[u]=col;
	for(int i=0;i<g[u].size();i++){
		int v=g[u][i];
		if(!vis[v]) DFS(v,sl,sz,(col+1)%3);
		else if((col+1)%3!=c[v]) sl=true;
	}
	for(int i=0;i<r[u].size();i++){
		int v=r[u][i];
		if(!vis[v]) DFS(v,sl,sz,(col+2)%3);
		else if((col+2)%3!=c[v]) sl=true;
	}
}
int main(){
	int n,m;
	long long ans=0;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) f[i]=i;
	for(int i=0;i<m;i++){
		scanf("%d%d",&a[i],&b[i]);
		out[a[i]]++;
		in[b[i]]++;
		g[a[i]].push_back(b[i]);
		r[b[i]].push_back(a[i]);
	}
	for(int i=1;i<=n;i++){
		if(in[i]&&out[i]&&!vis[i]){
			bool sl=false;
			int sz[3]={};
			DFS(i,sl,sz,0);
			if(sl){
				int tsz=sz[0]+sz[1]+sz[2];
				ans+=1LL*tsz*tsz;
			}
			else{
				ans+=1LL*sz[0]*sz[1]+1LL*sz[1]*sz[2]+1LL*sz[2]*sz[0];
			}
		}
	}
	for(int i=0;i<m;i++){
		if(!vis[a[i]]) ans++;
	}
	printf("%lld\n",ans);
	return 0;
}