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>
using namespace std;

typedef long long ll;
const int maxn=100010;
int n,m;
vector<int> g[maxn], gg[maxn];
int cnt, vis[maxn], flag, in[maxn], tot[5], t;
ll ans;

void dfs(int x,int bel){
	vis[x]=1; cnt++; in[x]=bel; tot[bel]++;
	for(int i=0;i<g[x].size();i++){
		t++; int v=g[x][i]; if(v==x) flag=1; if(vis[v] && in[v]!=(bel+1)%3) flag=1; if(vis[v]) continue;
		dfs(v, (bel+1)%3);
	}
	for(int i=0;i<gg[x].size();i++){
		int v=gg[x][i]; if(v==x) flag=1; if(vis[v] && in[v]!=(bel-1+3)%3) flag=1; if(vis[v]) continue;
		dfs(v, (bel-1+3)%3);
	}
}

int main(){

	scanf("%d%d",&n,&m);
	int x, y;
	for(int i=1;i<=m;i++) scanf("%d%d", &x, &y), g[x].push_back(y), gg[y].push_back(x);
	for(int i=1;i<=n;i++) if(!vis[i]){
		memset(tot,0,sizeof(tot));
		cnt=0; flag=0; t=0;
		dfs(i,0);
		// printf("i = %d cnt = %d t = %d\n", i, cnt, t);
		if(flag){ ans+=1ll*cnt*cnt; }
		else if(!tot[0] || !tot[1] || !tot[2]) ans+=t;
		else ans+=1ll*tot[0]*tot[1]+1ll*tot[1]*tot[2]+1ll*tot[2]*tot[0];
		// printf("ans = %d\n", ans);
	}
	printf("%lld\n", ans);
	return 0;
}