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<bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=(a),_ed=(b);i<=_ed;++i)
#define DREP(i,a,b) for(int i=(a),_ed=(b);i>=_ed;--i)
typedef long long ll;
inline int read(){
    register int x=0,f=1;register char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}
    while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}
    return f?x:-x;
}

const int N=1e5+5;
int n,m,num,head[N],ecnt,cnt[3],flg,col[N];
struct edge{int to,val,next;} e[N<<1];
inline void add(int u,int v){
    e[++num]=(edge){v,1,head[u]},head[u]=num;
    e[++num]=(edge){u,2,head[v]},head[v]=num;
}

inline void dfs(int u){
    ++cnt[col[u]];
    for(int i=head[u];i;i=e[i].next){
	if(e[i].val==1)++ecnt;
	int v=e[i].to;
	if(!~col[v])col[v]=(col[u]+e[i].val)%3,dfs(v);
	else if(col[v]!=(col[u]+e[i].val)%3)flg=1;
    }
}

int main(){
    //freopen("in.in","r",stdin);
    n=read(),m=read();
    REP(i,1,m){int u=read(),v=read();add(u,v);}
    memset(col,-1,sizeof col);
    ll ans=0;
    REP(i,1,n)if(!~col[i]){
	ecnt=cnt[0]=cnt[1]=cnt[2]=flg=0;
	col[i]=0;dfs(i);
	if(flg){ans+=1ll*(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);continue;}
	if(!cnt[0]||!cnt[1]||!cnt[2]){ans+=ecnt;continue;}
	ans+=1ll*cnt[0]*cnt[1];
	ans+=1ll*cnt[1]*cnt[2];
	ans+=1ll*cnt[2]*cnt[0];
    }
    printf("%lld\n",ans);
    return 0;
}
