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:
//by xiqiao
#include<bits/stdc++.h>
#define maxn 100005
using namespace std;
typedef long long ll;
vector <int> v[maxn];
vector <int> g[maxn];
int n,m,flag;
int vis[maxn];
ll ans;
int cnt[4],ge;
int next(int x){if(x==3)return 1;return x+1;}
int next2(int x){if(x==1)return 3;if(x==2)return 1;return 2;}
void dfs(int x,int c){
	cnt[c]++;vis[x]=c;
	for(int i=0;i<(int)v[x].size();i++){
		if(v[x][i]==x)flag=1;
		ge++;
		if(!vis[v[x][i]]){
			dfs(v[x][i],next(c));
		}
		else if(vis[v[x][i]]!=next(c)){
			flag=1;
		}
	}
	for(int i=0;i<(int)g[x].size();i++){
		ge++;
		if(!vis[g[x][i]]){
			dfs(g[x][i],next2(c));
		}
		else if(vis[g[x][i]]!=next2(c)){
			flag=1;
		}	
	}
}
int main(){
//	freopen("table.in","r",stdin);
	//freopen("table.out","w",stdout);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int x,y;scanf("%d%d",&x,&y);
		v[x].push_back(y);g[y].push_back(x);
	}
	for(int i=1;i<=n;i++){
		if(!vis[i]){
			flag=0;cnt[1]=0;cnt[2]=0;cnt[3]=0;ge=0;
			dfs(i,1);
			int sum=cnt[1]+cnt[2]+cnt[3];
			if(flag){
				ans+=1ll*sum*sum;
			}
			else if(cnt[1] && cnt[2] && cnt[3]){
				ans+=1ll*cnt[1]*cnt[2];
				ans+=1ll*cnt[2]*cnt[3];
				ans+=1ll*cnt[3]*cnt[1];
			}
			else ans+=ge/2;
		}
	}
	cout<<ans<<endl;
	return 0;
}