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
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define o 300011
using namespace std;
ll total,edgecnt,num[4],ans;
int n,m,ui,vi,tot;
int color[o],head[o],nxt[o],point[o],remain[o];
bool vis[o],cir,g[o];


void addedge(int x,int y){
	tot++;nxt[tot]=head[x];head[x]=tot;point[tot]=y;remain[tot]=1;
	tot++;nxt[tot]=head[y];head[y]=tot;point[tot]=x;remain[tot]=-1;
}

void dfs(int now){
	vis[now]=true;
	total++;
	num[color[now]]++;
	for(int tmp=head[now];tmp;tmp=nxt[tmp]){
		if(g[tmp])continue;
		g[tmp]=g[tmp^1]=true;
		edgecnt++;
		int v=point[tmp];
		if(vis[v]){
			if(color[v]!=(color[now]+remain[tmp]+3)%3) cir=true;
		}
		else{
			color[v]=(color[now]+remain[tmp]+3)%3;
			dfs(v);
		}
	}
}
void clear(){
	total=0;edgecnt=0;
	num[1]=num[2]=num[0]=0;
	cir=false;
}
int main(){
	tot=1;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d",&ui,&vi);
		addedge(ui,vi);
	}
	for(int i=1;i<=n;i++)
	if(!vis[i]){
		clear();
		dfs(i);
		if(cir){
			ans+=total*total;
			continue;
		}
		if(!num[0]||!num[1]||!num[2]){
			ans+=edgecnt;
			continue;
		}
		ans+=num[0]*num[1];
		ans+=num[0]*num[2];
		ans+=num[1]*num[2];
	}
	printf("%lld\n",ans);
}