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 <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define For(i,u) for(int i=hed[u],v=e[i].to;i;i=e[i].nex,v=e[i].to)
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n,m,u,v,vis[N],f[N],hed[N],ed,eg,sz,col,cn;ll ans;
struct edge{
    int to,nex,val;
    edge(int _=0,int __=0,int ___=0):to(_),nex(__),val(___){}
}e[N<<1];
inline void add(int u,int v,int w){e[++ed]=edge(v,hed[u],w);hed[u]=ed;}
void dfs(int u,int c){
    !f[c]&&++col,vis[u]=c,++f[c],++sz;
    For(i,u){
        if(e[i].val==1) ++eg;
        if(!(~vis[v])) dfs(v,(c+e[i].val)%3);
        else if(vis[v]!=(c+e[i].val)%3) cn=0;
    }
}
int main(){
    scanf("%d%d",&n,&m);
    rep(i,1,m) scanf("%d%d",&u,&v),add(u,v,1),add(v,u,2);
    memset(vis,-1,sizeof(vis));
    rep(i,1,n) if(!(~vis[i])){
        sz=eg=col=0;cn=1;f[0]=f[1]=f[2]=0;dfs(i,0);
        ans+=cn?(col==3?(1ll*f[0]*f[1]+1ll*f[1]*f[2]+1ll*f[2]*f[0]):eg):(1ll*sz*sz);
    }
    printf("%lld\n",ans);
    return 0;
}