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 N=100005;
int read(){
	int x=0;
	char ch=getchar();
	while (!isdigit(ch))
		ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x;
}
int n,m;
vector <int> e[N];
int vis[N];
LL ans=0;
int cnt[3],cnte=0;
bool dfs(int x,int c){
	cnt[vis[x]=c]++;
	bool res=1;
	for (auto t : e[x]){
		int y=t>>1,z=t&1;
		if (!z)
			cnte++;
		int _c=(c+(z?1:-1)+3)%3;
		if (~vis[y])
			res&=vis[y]==_c;
		else
			res&=dfs(y,_c);
	}
	return res;
}
LL solve(int x){
	cnte=0;
	memset(cnt,0,sizeof cnt);
	if (!dfs(x,0)){
		int s=cnt[0]+cnt[1]+cnt[2];
		return 1LL*s*s;
	}
	if (!cnt[0]||!cnt[1]||!cnt[2])
		return cnte;
	return 1LL*cnt[0]*cnt[1]+1LL*cnt[1]*cnt[2]+1LL*cnt[2]*cnt[0];
}
int main(){
	n=read(),m=read();
	for (int i=1;i<=n;i++)
		e[i].clear();
	for (int i=1;i<=m;i++){
		int a=read(),b=read();
		e[a].push_back(b<<1|1);
		e[b].push_back(a<<1);
	}
	memset(vis,-1,sizeof vis);
	for (int i=1;i<=n;i++)
		if (!~vis[i])
			ans+=solve(i);
	printf("%lld",ans);
	return 0;
}