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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

typedef long long ll;
const int N=5e5+9;

ll to[N<<1],nxt[N<<1],beg[N],tot;
ll n,m,c[N],w[N<<1];
ll ans,f[4],flag;

inline void add(int u,int v,int c)
{
	to[++tot]=v;
	nxt[tot]=beg[u];
	w[tot]=c;
	beg[u]=tot;
}

inline void dfs(int u)
{
	f[c[u]]++;
	for(int i=beg[u];i;i=nxt[i])
	{
		if(w[i]==1)f[3]++;
		if(!(~c[to[i]]))
		{
			c[to[i]]=(c[u]+w[i])%3;
			dfs(to[i]);
		}
		else if (c[to[i]]!=(c[u]+w[i])%3)
			flag=1;
	}
}

int main()
{
	if(fopen("vj.in","r"))
	{
		freopen("vj.in","r",stdin);
		freopen("vj.out","w",stdout);
	}

	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,2);
	}

	memset(c,-1,sizeof(c));
	for(int i=1;i<=n;i++)
		if(!(~c[i]))
		{
			f[c[i]=0]=f[1]=f[2]=f[3]=flag=0;
			dfs(i);
			if(flag)
			{
				ll cnt=f[0]+f[1]+f[2];
				ans+=cnt*cnt;
				goto nxt;
			}
			if((!f[0]) || (!f[1]) || (!f[2]))
			{
				ans+=f[3];
				goto nxt;
			}
			
			ans+=f[0]*f[1];
			ans+=f[1]*f[2];
			ans+=f[2]*f[0];

			nxt:;
		}

	printf("%lld\n",ans);
	return 0;
}