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>
template<typename T>inline void read(T &x)
{
	char c=x=0;
	for(c=getchar();!isdigit(c);c=getchar());
	for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48);
}
namespace fucker
{
	typedef long long ll;
	const int N=100100,M=N*2;
	int begin[N],next[M],to[M],w[M];
	int col[N];
	int n,m,e;
	ll ans;
	void add(int x,int y,int z=1,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		w[e]=z;
		if(k)add(y,x,2,0);
	}
	void initialize()
	{
		read(n),read(m);
		for(int u,v;m--;)
			read(u),read(v),add(u,v);
	}
	int cnt[3],tmp;
	bool dfs(int p)
	{
		cnt[col[p]]++;

		bool ret=1;
		for(int i=begin[p],q;i;i=next[i])
		{
			q=to[i],tmp++;

			if(col[q]>=0) {ret&=(col[q]==((col[p]+w[i])%3));continue;}
			col[q]=(col[p]+w[i])%3;
			ret&=dfs(q);
		}
		return ret;
	}
	void solve()
	{
		initialize();
		for(int i=1;i<=n;i++)col[i]=-1;

		ans=0;
		for(int i=1;i<=n;i++)
			if(col[i]<0)
			{
				tmp=cnt[0]=cnt[1]=cnt[2]=col[i]=0;
				if(dfs(i))
				{
					if(cnt[0] && cnt[1] && cnt[2])
						ans+=(ll)cnt[0]*cnt[1]+(ll)cnt[1]*cnt[2]+(ll)cnt[0]*cnt[2];
					else ans+=tmp/2;
				}
				else ans+=(ll)(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);
			}

		printf("%lld\n",ans);
	}
}
int main()
{
//	freopen("in","r",stdin);
//	freopen("out","w",stdout);
	fucker::solve();
	return 0;
}