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;
#define ll long long
const int N=100005;
int n,m,cc,ff,sv,se,hd[N],vis[N],f[3];ll ans;
struct E{int v,nxt,w;}e[N*2];
inline void add(int u,int v,int w){e[++cc]=(E){v,hd[u],w};hd[u]=cc;}
void dfs(int x,int c)
{
	vis[x]=c;f[c]++;sv++;
	for(int i=hd[x];i;i=e[i].nxt)
	{
		int y=e[i].v;se+=(e[i].w==1);
		if(vis[y]==-1)dfs(y,(c+e[i].w)%3);
		else if(vis[y]!=(c+e[i].w)%3)ff=0;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1,u,v;i<=m;i++)scanf("%d%d",&u,&v),add(u,v,1),add(v,u,2);
	memset(vis,-1,sizeof(vis));
	for(int i=1;i<=n;i++)if(vis[i]==-1)
	{
		ff=1;f[0]=f[1]=f[2]=0;sv=se=0;dfs(i,0);
		ans+=!ff?1ll*sv*sv:(!min(f[0],min(f[1],f[2]))?se:(1ll*f[0]*f[1]+1ll*f[1]*f[2]+1ll*f[2]*f[0]));
	}
	printf("%lld\n",ans);return 0;
}