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 pb push_back

using namespace std;

const int N=100005,fc[3]={2,0,1},bc[3]={1,2,0};
int n,m,col[N],V[3],num;
long long ans;
bool vis[N],ok;
vector<int> E1[N],E2[N];

void dfs(int x,int c){
	col[x]=c,V[c]++,vis[x]=true,num+=E1[x].size(); 
	for(int v:E1[x])
		if(vis[v])ok&=col[v]==bc[c];else dfs(v,bc[c]);
	for(int v:E2[x])
		if(vis[v])ok&=col[v]==fc[c];else dfs(v,fc[c]);
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1,u,v;i<=m;i++)scanf("%d%d",&u,&v),E1[u].pb(v),E2[v].pb(u);
	for(int i=1;i<=n;i++)if(!vis[i]){
		int&a=V[0],&b=V[1],&c=V[2];
		ok=true,a=b=c=num=0,dfs(i,0);
		if(!ok)
			ans+=(long long)(a+b+c)*(a+b+c);
		else if(!a||!b||!c)
			ans+=num;
		else
			ans+=(long long)a*b+(long long)b*c+(long long)c*a;
	}
	printf("%lld\n",ans);
	return 0;
}