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 <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n,m,k,t,cnt[3],c[maxn],edge;
bool vis[maxn],flag;
vector<pair<int,int>>e[maxn];
void dfs(int x,int y)
{
    cnt[y]++;
    c[x]=y;
    for(int i=0;i<e[x].size();i++)
    {
        int to=e[x][i].first,w=e[x][i].second;
        if(w==1)edge++;
        if(!vis[to])
        {
            vis[to]=true;
            dfs(to,(y+w)%3);
        }
        else if(c[to]!=(y+w)%3)flag=false;
    }
}
int main()
{
    int i,j;
    scanf("%d%d",&n,&m);
    while(m--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].push_back({y,1}),e[y].push_back({x,2});
    }
    long long ret=0;
    for(i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            vis[i]=true;
            cnt[0]=cnt[1]=cnt[2]=0;
            flag=true;
            edge=0;
            dfs(i,0);
            if(!flag)
            {
                ret+=1LL*(cnt[0]+cnt[1]+cnt[2])*(cnt[0]+cnt[1]+cnt[2]);
            }
            else if(!cnt[2]||!cnt[1]||!cnt[0])
            {
                ret+=edge;
            }
            else ret+=1LL*cnt[0]*cnt[1]+1LL*cnt[1]*cnt[2]+1LL*cnt[2]*cnt[0];
        }
    }
    printf("%lld\n",ret);
    return 0;
}
