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 <bits/stdc++.h>

using namespace std;
#define X first
#define Y second
typedef double db;
typedef long long ll;
typedef pair<int,int> P;
const int MAXN=1e5+10;
struct edge{int nxt,to,w;}e[MAXN<<2];
ll res=0,cnt[5];bool f;
int n,m,x,y,head[MAXN],col[MAXN],tot,ne,nd;

void add(int x,int y,int w)
{e[++tot]=(edge){head[x],y,w};head[x]=tot;}
void dfs(int x)
{
	cnt[col[x]]++;nd++;
	for(int i=head[x];i;i=e[i].nxt)
	{
		ne+=(e[i].w==1);
		if(col[e[i].to]==-1)
			col[e[i].to]=(col[x]+e[i].w)%3,dfs(e[i].to);
		else if(col[e[i].to]!=(col[x]+e[i].w)%3) f=false;
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	memset(col,-1,sizeof(col));
	for(int i=1;i<=m;i++)
		scanf("%d%d",&x,&y),add(x,y,1),add(y,x,2);
	for(int i=1;i<=n;i++)
	{
		if(~col[i]) continue;
		cnt[0]=cnt[1]=cnt[2]=0;
		col[i]=0;nd=ne=0;f=true;dfs(i);
		
//		assert(nd==cnt[0]+cnt[1]+cnt[2]);
		if(!f) res+=1ll*nd*nd;
		else if(!cnt[0]||!cnt[1]||!cnt[2]) res+=ne;
		else res+=cnt[0]*cnt[1]+cnt[1]*cnt[2]+cnt[2]*cnt[0];
	}
	printf("%lld",res);
	return 0;
}