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>
const int N=100005;
using namespace std;
struct edge{
	int to,next,v;
}e[N*2];
int head[N],tot=1;
void add(int x,int y,int v){
	e[++tot]=(edge){y,head[x],v};
	head[x]=tot;
}
int co[N],cnt[3];
int cnte,cntp,ok;
bool vis[N*2];
void dfs(int x,int c){
	if (co[x]!=-1){
		if (co[x]!=c)
			ok=0;
		return;
	}
	co[x]=c; cnt[c]++; cntp++;
	for (int i=head[x];i;i=e[i].next)
		if (!vis[i]&&!vis[i^1]){
			vis[i]=vis[i^1]=1; cnte++;
			dfs(e[i].to,(c+e[i].v)%3);
		}
}
int n,m;
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y,1); add(y,x,2);
	}
	memset(co,-1,sizeof(co));
	long long ans=0;
	for (int i=1;i<=n;i++)
		if (co[i]==-1){
			memset(cnt,0,sizeof(cnt));
			ok=1; cnte=0; cntp=0; dfs(i,0);
			if (!ok) ans+=1ll*cntp*cntp;
			else if (!cnt[0]||!cnt[1]||!cnt[2]) ans+=cnte;
			else ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[2]*cnt[0];
		}
	printf("%lld",ans);
}