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 <cctype>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

typedef long long LL;
int n,m,c[100100],cnt[3],cnte;
LL ans;
struct edge
{
    int v,w;
    edge *nxt;
} pool[200100],*tp=pool,*fst[100100];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()));
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

bool dfs(int x,int w)
{
    if (c[x]!=-1)
        return c[x]!=w;
    ++cnt[c[x]=w];
    bool flag=0;
    for (edge *i=fst[x]; i; ++cnte,i=i->nxt)
        flag|=dfs(i->v,(w+i->w)%3);
    return flag;
}

int main()
{
    scanf("%d%d",&n,&m);
    while (m--)
    {
        int u=getint(),v=getint();
        *tp=(edge){v,1,fst[u]},fst[u]=tp++;
        *tp=(edge){u,2,fst[v]},fst[v]=tp++;
    }
    memset(c,255,sizeof(c));
    repu(i,1,n)
        if (c[i]==-1)
        {
            cnt[0]=cnt[1]=cnt[2]=cnte=0;
            if (dfs(i,0))
            {
                ans+=LL(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);
                continue;
            }
            if (!cnt[0] || !cnt[1] || !cnt[2])
                ans+=cnte/2;
            else
                ans+=LL(cnt[0])*cnt[1]+LL(cnt[1])*cnt[2]+LL(cnt[2])*cnt[0];
        }
    printf("%lld\n",ans);
    return 0;
}