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
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(a,b,c) for (int a=b;a<=c;a++)
#define per(a,b,c) for (int a=b;a>=c;a--)
#define go(u) for (int o=ft[u],v;v=E[o].t;o=E[o].n)
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int,int> par;
const int N=100010;
struct edge{int t,l,n;}E[N<<2];
int n,m,h,tot,t,q[N],tp[N],vis[N],ft[N];
LL ans;
void add(int x,int y,int z){
	E[++tot]=(edge){y,z,ft[x]},ft[x]=tot;
}
LL solve(int u){
	LL s[4]={0,0,0,0},flag=0,s2=0;
	q[h=t=1]=u;
	vis[u]=1;
	while (h<=t){
		u=q[h++];
		s[3]++;
		s[tp[u]]++;
		go(u) { 
			++s2;if (!vis[v]){
			vis[v]=1;
			tp[v]=(tp[u]+E[o].l+3)%3;
			q[++t]=v;
		} else if (tp[v]!=(tp[u]+E[o].l+3)%3) flag=1;} 
	}
	if (flag) return s[3]*s[3];
	if (!s[0]||!s[1]||!s[2]) return s2/2;
	return s[0]*s[1]+s[1]*s[2]+s[2]*s[0];
}
int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,m){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y,1),add(y,x,-1);
	}
	rep(i,1,n) if (!vis[i]) ans+=solve(i);
	printf("%lld\n",ans);
	return 0;
}
