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<cstdio>
#include<cstring>

const int N=100005;

int n,m,te,h[N],nxt[N<<1],to[N<<1],w[N<<1],f[N],cnt[3],e,ok;
long long ans;

void link(int a,int b,int c)
{
	nxt[++te]=h[a];
	h[a]=te;
	to[te]=b;
	w[te]=c;
}

void dfs(int u)
{
	cnt[f[u]]++;
	for(int i=h[u],v;i;i=nxt[i])
	{
		++e;
		v=to[i];
		if(-1==f[v])
		{
			f[v]=(f[u]+w[i])%3;
			dfs(v);
		}
		else if((f[u]+w[i])%3!=f[v])
			ok=0;
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	while(m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		link(a,b,1),link(b,a,2);
	}
	memset(f,-1,sizeof(f));
	for(int i=1; i<=n; i++)
		if(-1==f[i])
		{
			ok=1;
			cnt[0]=cnt[1]=cnt[2]=0;
			e=0;
			f[i]=0;
			dfs(i);
			e/=2;
			if(!ok)
				ans+=1ll*(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);
			else if(cnt[0]&&cnt[1]&&cnt[2])
				ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[2]*cnt[0];
			else
				ans+=e;
		}
	printf("%lld",ans);
	return 0;
}