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 rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long

using namespace std;

const int N=1e5+10;
int n,m,x,y,cnt,head[N],col[N],sz,flag; ll t[3],ans;
struct edge{int to,nxt,w;}e[N<<1];

void adde(int x,int y,int w){
	e[++cnt].to=y; e[cnt].nxt=head[x]; head[x]=cnt; e[cnt].w=w;
}

void dfs(int u,int c){
	t[col[u]=c]++;
	for (int i=head[u],v;i;i=e[i].nxt){
		v=e[i].to,sz+=e[i].w==1;
		if (col[v]!=-1) flag|=col[v]!=(c+e[i].w)%3;
		else dfs(v,(c+e[i].w)%3);
	}
}

int main(){
	scanf("%d%d",&n,&m);
	rep (i,1,m) scanf("%d%d",&x,&y),adde(x,y,1),adde(y,x,2);
	memset(col,-1,sizeof(col));
	rep (i,1,n) if (col[i]==-1){
		flag=sz=t[0]=t[1]=t[2]=0,dfs(i,0);
		if (flag) ans+=(t[0]+t[1]+t[2])*(t[0]+t[1]+t[2]);
		else if (!t[0]||!t[1]||!t[2]) ans+=sz;
		else ans+=t[0]*t[1]+t[1]*t[2]+t[2]*t[0];
	}
	printf("%lld\n",ans);
	return 0;
}