Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define N 100009
#define M 200009
using namespace std;

int n,m,sz,eg,tot,num[3],fst[N],pnt[M],len[M],nxt[M],vis[N];
bool flag;
void add(int x,int y,int z){
	pnt[++tot]=y; len[tot]=z; nxt[tot]=fst[x]; fst[x]=tot;
}
void dfs(int x,int t){
	if (vis[x]!=-1){
		if (vis[x]!=t) flag=1;
		return;
	}
	sz++; vis[x]=t; num[t]++;
	int i,y;
	for (i=fst[x]; i; i=nxt[i]){
		y=pnt[i];
		if (len[i]==1) eg++;
		dfs(y,(t+len[i])%3);
	}
}
int main(){
	scanf("%d%d",&n,&m);
	int i,x,y;
	for (i=1; i<=m; i++){
		scanf("%d%d",&x,&y);
		add(x,y,1); add(y,x,2);
	}
	memset(vis,-1,sizeof(vis));
	ll ans=0;
	for (i=1; i<=n; i++) if (vis[i]==-1){
		sz=eg=0; flag=0;
		memset(num,0,sizeof(num));
		dfs(i,0);
		if (flag) ans+=(ll)sz*sz; else
		if (!num[0] || !num[1] || !num[2]) ans+=eg; else
		ans+=(ll)num[0]*num[1]+(ll)num[1]*num[2]+(ll)num[2]*num[0];
	}
	printf("%lld\n",ans);
	return 0;
}