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;

typedef long long ll;
const int maxn=1e5+10;
int n,m,fa[maxn],vis[maxn],size[3],cnt;
ll ans;
vector<int> g[maxn],gg[maxn];
bool flag;

void dfs(int pos,int c){
	if(~vis[pos]){
		if(vis[pos]!=c)
			flag=true;
		return;
	}
	vis[pos]=c;
	++size[c];
	cnt+=g[pos].size();
	for(int i=0;i<g[pos].size();++i)
		dfs(g[pos][i],(c+1)%3);
	for(int i=0;i<gg[pos].size();++i)
		dfs(gg[pos][i],(c+2)%3);
}

int main(){
	cin>>n>>m;
	for(int i=1,u,v;i<=m;++i){
		cin>>u>>v;
		g[u].push_back(v);
		gg[v].push_back(u);
	}
	memset(vis,-1,sizeof(vis));
	for(int i=1;i<=n;++i)
		if(vis[i]==-1){
			flag=false;
			size[0]=size[1]=size[2]=0;
			cnt=0;
			dfs(i,0);
			if(flag)
				ans+=(ll)(size[0]+size[1]+size[2])*(size[0]+size[1]+size[2]);
			else if(size[1]&&size[2])
				ans+=(ll)size[0]*size[1]+(ll)size[1]*size[2]+(ll)size[2]*size[0];
			else
				ans+=cnt;
		}
	printf("%lld\n",ans);
	return 0;
}