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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;
#define pb push_back
const int maxn = 200005;

int ehead[maxn],ecnt;
struct edge {
	int u,v,w,next;
}edg[maxn];
void add(int u,int v) {
	edg[++ecnt]=(edge){u,v,1,ehead[u]};
	ehead[u]=ecnt;
	edg[++ecnt]=(edge){v,u,2,ehead[v]};
	ehead[v]=ecnt;
}
int n,m,col[maxn],cnt[3],siz,tag;LL ans;
LL sqr(LL x) {return x*x;}

void dfs(int u)
{
	cnt[col[u]]++;
	for (int j=ehead[u];j;j=edg[j].next) {
		int v=edg[j].v;++siz;
		if (col[v]<0) col[v]=(col[u]+edg[j].w)%3,dfs(v);
		else if (col[v]!=(col[u]+edg[j].w)%3) tag=false;
	}
}
int main()
{
	#ifdef Amberframe
		freopen("agc006f.in","r",stdin);
		freopen("agc006f.out","w",stdout);
	#endif
	scanf("%d %d",&n,&m);
	for (int u,v,i=1;i<=m;i++)
		scanf("%d %d",&u,&v),add(u,v);
	memset(col,-1,sizeof col);
	for (int i=1;i<=n;i++)
	if (col[i]<0) {
		cnt[0]=cnt[1]=cnt[2]=siz=0;
		col[i]=0;tag=true;dfs(i);
		if (!tag) ans+=sqr(cnt[0]+cnt[1]+cnt[2]);
		else if (cnt[0]&&cnt[1]&&cnt[2])
			ans+=(LL)cnt[0]*(cnt[1]+cnt[2])+(LL)cnt[1]*cnt[2];
		else ans+=siz>>1;
	}
	printf("%lld\n",ans);
	return 0;
}