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<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
vector<pair<int,int> >g[110000];
bool th=true;
int v[110000];
int num[110000];
int cnt[3];
int tc;
void dfs(int a,int b){
 
	num[a]=b;
	cnt[b]++;
	for(int i=0;i<g[a].size();i++){
		if(g[a][i].second==1)tc++;
		if(!v[g[a][i].first]){
			v[g[a][i].first]=1;
			dfs(g[a][i].first,(b+g[a][i].second)%3);
		}else{
			if(num[g[a][i].first]!=(b+g[a][i].second)%3)th=false;
		}
	}
}
 
int main(){
	int a,b;scanf("%d%d",&a,&b);
	for(int i=0;i<b;i++){
		int p,q;scanf("%d%d",&p,&q);
		p--;q--;
		g[p].push_back(make_pair(q,1));
		g[q].push_back(make_pair(p,2));
	}
	long long ret=0;
	for(int i=0;i<a;i++){
		if(!v[i]){
			for(int j=0;j<3;j++)cnt[j]=0;
			th=true;
			v[i]=1;
			tc=0;
			dfs(i,0);
			if(th){
				if(cnt[0]&&cnt[1]&&cnt[2])ret+=(long long)cnt[0]*cnt[1]+(long long)cnt[1]*cnt[2]+(long long)cnt[2]*cnt[0];
				else ret+=tc;
			}else{
				int tot=cnt[0]+cnt[1]+cnt[2];
				ret+=(long long)tot*tot;
			}
		}
	}
	printf("%lld\n",ret);
}