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<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 100000
using namespace std;
typedef long long LL;
struct node
{
	int to,w;
	node *nxt;
}edges[MAXN*4+5];
node *ncnt,*Adj[MAXN+5];
int N,M;
int col[MAXN+5],cnt[3],edcnt=0;
bool legal=true;
void Init()
{
	ncnt=&edges[0];
	memset(col,-1,sizeof(col));
}
void AddEdge(int u,int v,int w)
{
	node *p=++ncnt;
	p->to=v;
	p->w=w;
	p->nxt=Adj[u];
	Adj[u]=p;
}
void DFS(int u)
{
	cnt[col[u]]++;
	for(node *p=Adj[u];p!=NULL;p=p->nxt)
	{
		int v=p->to;
		int w=p->w;
		if(w==1)
			edcnt++;
		if(col[v]==-1)
		{
			col[v]=(col[u]+w)%3;
			DFS(v);
		}
		else if(col[v]!=(col[u]+w)%3)
			legal=false;
	}
}
int main()
{
//	freopen("flower.in","r",stdin);
//	freopen("flower.out","w",stdout);
	Init();
	scanf("%d %d",&N,&M);
	int u,v;
	for(int i=1;i<=M;i++)
	{
		scanf("%d %d",&u,&v);
		AddEdge(u,v,1);
		AddEdge(v,u,2);
	}
	LL ans=0;
	for(int i=1;i<=N;i++)
		if(col[i]==-1)
		{
			edcnt=0;
			legal=true;
			memset(cnt,0,sizeof(cnt));
			col[i]=0;
			DFS(i);
			if(legal==false)
				ans+=(1LL*cnt[0]+1LL*cnt[1]+1LL*cnt[2])*(1LL*cnt[0]+1LL*cnt[1]+1LL*cnt[2]);
			else if(cnt[0]==0||cnt[1]==0||cnt[2]==0)
				ans+=1LL*edcnt;
			else if(legal==true)
				ans+=1LL*cnt[0]*cnt[1]+1LL*cnt[1]*cnt[2]+1LL*cnt[0]*cnt[2];
		}
	printf("%lld\n",ans);
	return 0;
}