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;
const int N=100100;
typedef long long ll;
int n,m,sz[3],vis[N],all_sz,cnt,flag;
vector<pair<int,int> >g[N];
void dfs(int x,int y){
	vis[x]=y;sz[y]++;all_sz++;
	for(auto i:g[x]){
		++cnt;
		int u=(y+i.second)%3;
		if(vis[i.first]==-1)dfs(i.first,u);
		else if(vis[i.first]!=u)flag=0;
	}
}
ll ans;
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	memset(vis,-1,sizeof vis);
	for(int i=1;i<=m;++i){
		int x,y;cin>>x>>y;
		g[x].emplace_back(y,1);
		g[y].emplace_back(x,2);
	}
	for(int i=1;i<=n;++i)if(vis[i]==-1){
		flag=1;memset(sz,0,sizeof sz);all_sz=0;cnt=0;
		dfs(i,0);
		if(!flag)ans+=(ll)all_sz*all_sz;
		else if(sz[0]&&sz[1]&&sz[2])ans+=(ll)sz[0]*sz[1]+(ll)sz[1]*sz[2]+(ll)sz[2]*sz[0];
		else ans+=cnt/2;
	}
	cout<<ans<<'\n';
	return 0;
}