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>

using namespace std;

typedef long long LL;
#define N 120000

LL n,m,cnt[3],E,col[N][3],ans;
bool vis[N];
vector<LL> g[N],h[N];

void dfs(LL u,LL c){
	if (!vis[u]){vis[u]=1; E+=g[u].size();}
	if (col[u][c]) return;
	col[u][c]=1; ++cnt[c];
	for (LL v:g[u]) dfs(v,(c+1)%3);
	for (LL v:h[u]) dfs(v,(c+2)%3);
}

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=m;++i){
		LL x,y; scanf("%lld%lld",&x,&y);
		g[x].push_back(y);
		h[y].push_back(x);
	}
	for (LL i=1;i<=n;++i)
		if (!vis[i]){
			cnt[0]=cnt[1]=cnt[2]=E=0;
			dfs(i,0);
			if (!cnt[1]||!cnt[2]) ans+=E;
			else{
				if (col[i][0]&&col[i][1]) ans+=cnt[0]*cnt[0];
				else ans+=cnt[0]*cnt[2]+cnt[1]*cnt[0]+cnt[2]*cnt[1];
			}
		}
	printf("%lld\n",ans);
	
	return 0;
}