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 <cstdio>

const int N=100005;

int fst[N],a[N],cnt[3],tot=0,num;
bool flag;

struct Edge{
	int to,v,nxt;
}e[N*2];

void addedge(int x,int y){
	e[++tot].to=y;e[tot].v=1;e[tot].nxt=fst[x];fst[x]=tot;
	e[++tot].to=x;e[tot].v=-1;e[tot].nxt=fst[y];fst[y]=tot;
}

void dfs(int x){
	cnt[a[x]]++;
	for (int i=fst[x];i;i=e[i].nxt){
		int y=e[i].to,z=(a[x]+e[i].v+3)%3;
		num++;
		if (a[y]==-1){
			a[y]=z;
			dfs(y);
		}
		else if (a[y]!=z) flag=1;
	}
}

int main(){
	int n,m;
	scanf("%d%d\n",&n,&m);
	for (int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d\n",&x,&y);
		addedge(x,y);
	}
	for (int i=1;i<=n;i++) a[i]=-1;
	long long ans=0;
	for (int i=1;i<=n;i++)
	if (a[i]==-1){
		for (int j=0;j<3;j++) cnt[j]=0;
		flag=0;a[i]=num=0;dfs(i);
		if (flag) ans+=1ll*(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);
		else if (!cnt[0] || !cnt[1] || !cnt[2]) ans+=num/2;
		else for (int i=0;i<3;i++) ans+=1ll*cnt[i]*cnt[(i+1)%3];
	}
	printf("%lld\n",ans);
}