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 ll long long
using namespace std;

int head[100005],cnt,vis[100005],color[100005],col,n,m;

ll ans,sum[3],ttot,ltot;

struct edge{
	int to,nxt,w;
}e[200005];

void add(int u,int v,int w){
	e[++cnt]=(edge){v,head[u],w};
	head[u]=cnt;
}

void dfs(int u){
	vis[u]=1;ttot++;
	for(int i=head[u];i;i=e[i].nxt){
		int v=e[i].to,w=e[i].w;
		if(w>0) ltot++;
		if(vis[v]){
			if(color[v]!=(color[u]+w+3)%3) col=-1;
			continue;
		}
		color[v]=(color[u]+w+3)%3;
		if(col!=-1&&sum[color[v]]==0) col++;
		sum[color[v]]++;
		dfs(v);
	}
}

signed main(){
	scanf("%d%d",&n,&m);
	for(int i=1,u,v;i<=m;i++){
		scanf("%d%d",&u,&v);
		add(u,v,1);add(v,u,-1);
	}
	for(int i=1;i<=n;i++) if(!vis[i]){
		ttot=ltot=0;col=1;
		sum[0]=1;sum[1]=sum[2]=0;
		color[i]=0;
		dfs(i);
		if(col==-1) ans+=ttot*ttot;
		else if(col==3) ans+=sum[0]*sum[1]+sum[1]*sum[2]+sum[2]*sum[0];
		else ans+=ltot;
	}
	printf("%lld",ans);
}