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<algorithm>
#include<cctype>
#include<vector>
using namespace std;
typedef int mmp;
#define vd void
#define rg register
#define il inline
#define sta static
il int gi(){
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return x*f;
}
const int maxn=1e5+1,maxm=maxn<<1;
int fir[maxn],dis[maxm],nxt[maxm],w[maxm],id;
il vd link(int a,int b,int c){nxt[++id]=fir[a],fir[a]=id,dis[id]=b,w[id]=c;}
vector<int>S;
int col[maxn],faq[maxn],tot,yes[maxm];
il int fuck(int x){
	while(x<1)x+=3;
	while(x>3)x-=3;
	return x;
}
il vd dfs(int x,int c){
	if(col[x]){
		if(c!=col[x])faq[x]=1;
		return;
	}
	S.push_back(x);col[x]=c;
	for(rg int i=fir[x];i;i=nxt[i]){
		if(!yes[i])++tot,yes[i]=1;
		dfs(dis[i],fuck(c+w[i]));
	}
}
mmp main(){
	int n=gi(),m=gi(),x,y;
	while(m--){
		x=gi(),y=gi();
		link(x,y,1);
		link(y,x,-1);
	}
	long long ans=0;
	for(rg int i=1;i<=n;++i)
		if(!col[i]){
			tot=0;S.clear();
			dfs(i,1);
			int a[4]={0},d=0;
			for(rg int i=0;i<S.size();++i){
				if(faq[S[i]]){d=1;break;}
				++a[col[S[i]]];
			}
			if(d)ans+=S.size()*S.size();
			else if(a[1]==0||a[2]==0||a[3]==0)ans+=tot/2;
			else ans+=1ll*a[1]*a[2]+1ll*a[2]*a[3]+1ll*a[3]*a[1];
		}
	printf("%lld\n",ans);
	return 0;
}
