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
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=200005;
typedef long long ll;
int n,m,V,E,cnt[3],col[N];
bool vis[N],vise[N*2],colored;
ll ans;
struct node{
    int u,v,w,nxt;
}edge[N*2];
int head[N],mcnt=1;
void add_edge(int u,int v,int w){
    mcnt++;
    edge[mcnt].u=u;
    edge[mcnt].v=v;
    edge[mcnt].w=w;
    edge[mcnt].nxt=head[u];
    head[u]=mcnt;
}
void dfs(int u)
{
	V++; vis[u]=true;
	cnt[col[u]]++;
	for(int i=head[u];i;i=edge[i].nxt){
        if(vise[i|1])
            continue ;
        vise[i|1]=1,E++;
        int v=edge[i].v,w=edge[i].w;
        if(!vis[v]){
            col[v]=(col[u]+w+3)%3;
            dfs(v);
        }
		else if((col[u]+w+3)%3!=col[v])
            colored=true;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add_edge(x,y,1);
		add_edge(y,x,-1);
	}
	for(int i=1;i<=n;i++)
		if(!vis[i])
		{
			V=E=cnt[0]=cnt[1]=cnt[2]=colored=0;
			dfs(i);
			if(colored)
                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);
}
