Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int maxn=100010,mod=998244353;
#define MP make_pair
#define PB push_back
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline ll read(){
	char ch=getchar();ll x=0,f=0;
	while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
	while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
	return f?-x:x;
}
inline int qmo(int x){return x+(x>>31?mod:0);}
int n,m,el,head[maxn],to[maxn*2],nxt[maxn*2],w[maxn*2],col[maxn],cnt[3],ecnt;
ll ans;
bool flag;
inline void add(int u,int v,int w_){
	to[++el]=v;nxt[el]=head[u];head[u]=el;w[el]=w_;
}
void dfs(int u){
	cnt[col[u]]++;
	for(int i=head[u];i;i=nxt[i]){
		int v=to[i];
		ecnt++;
		if(col[v]==-1) col[v]=(col[u]+w[i]+3)%3,dfs(v);
		else if(col[v]!=(col[u]+w[i]+3)%3) flag=false;
	}
}
int main(){
	n=read();m=read();
	FOR(i,1,m){
		int u=read(),v=read();
		add(u,v,1);add(v,u,-1);
	}
	MEM(col,-1);
	FOR(i,1,n) if(col[i]==-1){
		col[i]=0;
		flag=true;
		cnt[0]=cnt[1]=cnt[2]=ecnt=0;
		dfs(i);
		if(flag){
			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+=ecnt/2; 
		}
		else{
			int tot=cnt[0]+cnt[1]+cnt[2];
			ans+=1ll*tot*tot;
		}
	}
	printf("%lld\n",ans);
}