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
using namespace std;
const int N=1e5+10;
int n,m,cnt,num1,num2,chk;
int head[N],vis[N],col[N],to[N<<1],nxt[N<<1],val[N<<1],stk[3];
ll ans;
inline void add(int u,int v,int w){
	to[++cnt]=v,nxt[cnt]=head[u],val[cnt]=w,head[u]=cnt;
}
inline int tp(int x){ x=(x%3+3)%3; return x; }
void dfs(int u,int c){
	vis[u]=1,col[u]=c,stk[col[u]]++,num1++;
	for(int i=head[u];i;i=nxt[i]){
		int v=to[i];
		num2+=(val[i]==1);
		if(!vis[v]) dfs(v,tp(col[u]+val[i]));
		else if(col[v]!=tp(col[u]+val[i])) chk=1;
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) { int u,v; scanf("%d%d",&u,&v); add(u,v,1),add(v,u,-1); }
	for(int i=1;i<=n;i++){
		if(vis[i]) continue; 
		num1=0,num2=0,chk=0; stk[0]=stk[1]=stk[2]=0; 
		dfs(i,0);
		if(chk) ans+=1ll*num1*num1;
		else if(stk[0] && stk[1] && stk[2]) ans+=1ll*stk[0]*stk[1]+1ll*stk[1]*stk[2]+1ll*stk[2]*stk[0];
		else ans+=num2;
	}
	printf("%lld",ans);
}