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<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,m,tot,cnt1,cnt2,clr[100010],cnt[3],vis[100010],head[100010],nxt[200010],val[200010],to[200010];
ll ans=0;
void add_edge(int u,int v,int w){
	nxt[++tot]=head[u];
	to[tot]=v;
	val[tot]=w;
	head[u]=tot;
	return;
}
bool dfs(int u){
	bool ret=1;
	cnt[clr[u]]++;
	vis[u]=1;
	cnt1++;
	for(int i=head[u];~i;i=nxt[i]){
		int v=to[i],w=val[i],tmp=clr[u]+w;
		if(tmp>=3)
			tmp-=3;
		if(tmp<0)
			tmp+=3;
		cnt2+=(w>0);
		if(!vis[v]){
			clr[v]=tmp;
			if(!dfs(v))
				ret=0;
		}
		else
			if(clr[v]!=tmp)
				ret=0;
	}
	return ret;
}
char buffer[10000010],*hd;
char Getchar(){
	return *hd++;
}
int rd(){
	int x=0;
	char c;
	do c=Getchar();
	while(!isdigit(c));
	do{
		x=(x<<1)+(x<<3)+(c^48);
		c=Getchar();
	}while(isdigit(c));
	return x;
}
int main(){
	fread(buffer,1,10000000,stdin);
	hd=buffer;
	memset(head,-1,sizeof(head));
	n=rd();m=rd();
	for(int i=1;i<=m;i++){
		int u=rd(),v=rd();
		add_edge(u,v,1);
		add_edge(v,u,-1);
	}
	for(int i=1;i<=n;i++){
		if(vis[i])
			continue;
		cnt[0]=cnt[1]=cnt[2]=cnt1=cnt2=0;
		if(!dfs(i))
			ans+=1ll*cnt1*cnt1;
		else if(cnt[0]&&cnt[1]&&cnt[2])
			ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[0]*cnt[2]+1ll*cnt[1]*cnt[2];
		else ans+=cnt2;
	}
	printf("%lld",ans);
	return 0;
}