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;

typedef long long ll;
typedef pair<int, int> pii;

const int MN = 100010;

int N, M;
vector<pii> adj[MN];
int vis[MN], col[MN], cnt[3], ecnt;
bool ok;

void dfs(int u) {
    vis[u] = 1;
    cnt[ col[u] ]++;

    for(int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i].first;
        int w = adj[u][i].second;
        ecnt++;

        if(vis[v]) {
            if(col[v] != (col[u] + w) % 3) ok = false;
            continue;
        }
        col[v] = (col[u] + w) % 3;
        dfs(v);
    }
}

int main() {
    scanf("%d %d", &N, &M);

    for(int i = 0; i < M; i++) {
        int u, v; scanf("%d %d", &u, &v);
        u--; v--;

        adj[u].push_back(pii(v, 1));
        adj[v].push_back(pii(u, 2));
    }

    ll ans = 0;
    for(int i = 0; i < N; i++) if(!vis[i]) {
        memset(cnt, 0, sizeof(cnt));
        ecnt = 0;
        ok = true;
        dfs(i);

        if(ok) {
            if(!cnt[0] || !cnt[1] || !cnt[2]) {
                ans += ecnt / 2;
                continue;
            }

            for(int j = 0; j < 3; j++) {
                ans += 1LL * cnt[j] * cnt[(j + 1) % 3];
            }
        }
        else ans += 1LL * (cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
    }
    printf("%lld", ans);
}