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;
int n,m,x,y,col[100005],cnt[5],ok;long long ans;vector<pair<int,int> >e[100005];
void dfs(int x)
{
	cnt[col[x]]++;
	for(int i=0;i<e[x].size();i++)
	{
		if(e[x][i].second==1) cnt[3]++;
		if(col[e[x][i].first]==-1) col[e[x][i].first]=(col[x]+e[x][i].second)%3,dfs(e[x][i].first);
		else if((col[x]+e[x][i].second)%3!=col[e[x][i].first]) ok=1;
	}
}
int main()
{
	scanf("%d%d",&n,&m);memset(col,-1,sizeof(col));
	for(int i=1;i<=m;i++) scanf("%d%d",&x,&y),e[x].push_back(make_pair(y,1)),e[y].push_back(make_pair(x,2));
	for(int i=1;i<=n;i++) if(col[i]==-1)
	{
		memset(cnt,0,sizeof(cnt));ok=col[i]=0;dfs(i);
		if(ok){ans+=1ll*(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);continue;}
		else if((!cnt[0])||(!cnt[1])||(!cnt[2])){ans+=1ll*cnt[3];continue;}
		else ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[2]*cnt[0];
	}
	printf("%lld\n",ans);
}