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>
#define maxn 100005
#define LL long long
int n,m,V,E,cnt[3],col[maxn];
bool vis[maxn],vise[maxn],fg;
LL ans;
struct node { int v,w,i; node *nxt; } edge[maxn*2],*head[maxn],*ncnt;
void addedge(int u,int v,int w,int i)
{
	ncnt++;
	ncnt->v=v,ncnt->w=w,ncnt->nxt=head[u],ncnt->i=i;
	head[u]=ncnt;
}
void dfs(int u)
{
	V++; vis[u]=true;
	cnt[col[u]]++;
	for(node *p=head[u];p;p=p->nxt)
	{
		if(vise[p->i/2]) continue;
		vise[p->i/2]=true,E++;
		int v=p->v;
		if(!vis[v]) { col[v]=(col[u]+p->w+3)%3; dfs(v); }
		else if((col[u]+p->w+3)%3!=col[v]) fg=true;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	ncnt=&edge[0];
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		addedge(x,y,1,i*2); addedge(y,x,-1,i*2+1);
	}
	for(int i=1;i<=n;i++)
		if(!vis[i])
		{
			V=E=cnt[0]=cnt[1]=cnt[2]=fg=0;
			dfs(i);
			if(fg) ans+=1ll*V*V;
			else if(!cnt[0]||!cnt[1]||!cnt[2]) ans+=1ll*E;
			else ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[0]*cnt[2];
		}
	printf("%lld\n",ans);
}