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<stdio.h>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn = 2e5+5;
int n,m;
int la[maxn],nt[maxn],en[maxn],owo,len[maxn];
void adg(int x,int y,int z) {
	en[++owo]=y; nt[owo]=la[x]; la[x]=owo; len[owo]=z;
}
bool vis[maxn];
int col[maxn];
int CNT,BJ;int woc,orz[3];
void colorit(int x) {
	BJ++;
	vis[x] = 1; 
	orz[col[x]]++;
	for(int it=la[x];it;it=nt[it]) {
		int y = en[it];
		if(len[it]==1) CNT++;
		if(vis[y]) {
			if(col[y]!=(col[x]+len[it]+3)%3) woc = 1;
		} else {
			col[y] = (col[x]+len[it]+3)%3;
			colorit(y);
		}
	}
}
long long ans;
int main() {
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) {
		int x,y; scanf("%d%d",&x,&y);
		adg(x,y,1); adg(y,x,-1);
	}
	for(int i=1;i<=n;i++) {
		if(!vis[i]) {
			CNT = woc = orz[0] = orz[1] = orz[2] = 0; BJ=  0;
			colorit(i);
			if(woc) ans += 1ll*BJ*BJ;
			else if((!orz[0])||(!orz[1])||(!orz[2])) ans += CNT;
			else ans += 1ll*orz[0]*orz[1] + 1ll*orz[1]*orz[2] + 1ll*orz[2]*orz[0];
		}
	}
	printf("%lld",ans);
}