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>
#define ll long long
#define maxn 100010*5
using namespace std;

bool flag;

int n,m,cnt,head[maxn],col[maxn];

ll sum[5],ans;

struct E{
	int to,nxt,w;
}edge[maxn<<1];

void read(int &x){
	char s=getchar();x=0;
	while(s<'0'||s>'9') s=getchar();
	while(s>='0'&&s<='9') x=(x<<3)+(x<<1)+s-48,s=getchar();
}

inline void add(int u,int v,int w){edge[++cnt]=(E){v,head[u],w};head[u]=cnt;}

void dfs(int u){
	sum[col[u]]++;
	for(int i=head[u];i;i=edge[i].nxt){
		if(edge[i].w==1) sum[3]++;
		if(col[edge[i].to]==-1){
			col[edge[i].to]=(col[u]+edge[i].w)%3;
			dfs(edge[i].to);
		}
		else if(col[edge[i].to]!=(col[u]+edge[i].w)%3) flag=true;
	}
}

int main(){
	read(n);read(m);
	for(int i=1,x,y;i<=m;i++){
		read(x);read(y);
		add(x,y,1);
		add(y,x,2);
	}
	memset(col,-1,sizeof(col));
	for(int i=1;i<=n;i++){
		if(col[i]==-1){
			memset(sum,0,sizeof(sum));
			flag=col[i]=0;
			dfs(i);
			if(flag){
				ans+=(sum[0]+sum[1]+sum[2])*(sum[0]+sum[1]+sum[2]);
				continue;
			}
			if(sum[0]==0||sum[1]==0||sum[2]==0){
				ans+=sum[3];
				continue;
			}
			ans+=sum[0]*sum[1]+sum[1]*sum[2]+sum[2]*sum[0];
		}
	}
	printf("%lld",ans);
	return 0;
}