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;
#define int long long
const int N=100005;
int n,m,x,y,cnt[3],cntn,cntm,flag;
std::vector<int>e[N],er[N];
int col[N],ans=0;
void dfs(int x,int c){
	if (col[x]!=-1){
		if (col[x]!=c)flag=1;
		return;
	}
	cntn++;
	col[x]=c;
	cnt[col[x]]++;
	cntm+=e[x].size();
	for (int i=0;i<e[x].size();i++)dfs(e[x][i],(c+1)%3);
	for (int i=0;i<er[x].size();i++)dfs(er[x][i],(c+2)%3);
}
signed main(){
	scanf("%lld%lld",&n,&m);
	while (m--){
		scanf("%lld%lld",&x,&y);
		e[x].push_back(y);
		er[y].push_back(x);
	}
	memset(col,-1,sizeof col);
	for (int i=1;i<=n;i++)
		if (col[i]==-1){
			cnt[0]=cnt[1]=cnt[2]=0;
			cntn=cntm=flag=0;
			dfs(i,1);
			if (flag)ans+=cntn*cntn;
			else if (cnt[0]&&cnt[1]&&cnt[2])
				ans+=cnt[0]*cnt[1]+cnt[1]*cnt[2]+cnt[2]*cnt[0];
			else ans+=cntm;
		}
	printf("%lld\n",ans);
	return 0;
}