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
#include<bits/stdc++.h>
using namespace std;

vector<pair<int,int> > e[100001];
int vis[100001];
vector<int> w;
bool dfs(int x){
	bool res=false;
	w.emplace_back(x);
	for(auto it:e[x]){
		if(vis[it.first]!=-1){
			int y;
			if(it.second)
				y=(vis[x]+2)%3;
			else
				y=(vis[x]+1)%3;
			if(y!=vis[it.first])
				res=true;
		}
		else{
			if(it.second)
				vis[it.first]=(vis[x]+2)%3;
			else
				vis[it.first]=(vis[x]+1)%3;
			res|=dfs(it.first);
		}
	}
	return res;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	for(int i=1,u,v;i<=m;i++){
		cin>>u>>v;
		e[u].emplace_back(v,0);
		e[v].emplace_back(u,1);
	}
	for(int i=1;i<=n;i++)
		vis[i]=-1;
	long long int ans=0;
	for(int i=1;i<=n;i++)
		if(vis[i]==-1){
			w.clear();
			vis[i]=0;
			bool type=dfs(i);
			if(type)
				ans+=1LL*w.size()*w.size();
			else{
				int ww[3]={0,0,0};
				for(int it:w)
					ww[vis[it]]++;
				if(!ww[0]||!ww[1]||!ww[2]){
					long long int cnt=0;
					for(int it:w)
						cnt+=e[it].size();
					ans+=cnt/2;
				}
				else
					ans+=1LL*ww[0]*ww[1]+1LL*ww[1]*ww[2]+1LL*ww[2]*ww[0];
			}
		}
	cout<<ans<<endl;
}
