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>
#define maxn 100086

using namespace std;

int n, m;
int x, y;
vector<pair<int, int> > v[maxn];
int col[maxn];
long long ans;
int sum;
int cnt[3];
bool tag;


void dfs(int i){
	cnt[col[i]]++, sum += v[i].size();
	for(int j = 0;j < v[i].size();j++){
		int to = v[i][j].first, val = v[i][j].second;
		if(col[to] == -1) col[to] = (col[i] + val + 3) % 3, dfs(to);
		else if((col[i] + val + 3) % 3 != col[to]) tag = true;
	}
}

int main(){
	scanf("%d%d", &n, &m);
	while(m--){
		scanf("%d%d", &x, &y);
		v[x].push_back({y, 1}), v[y].push_back({x, -1});
	}
	memset(col, -1, sizeof(col));
	for(int i = 1;i <= n;i++){
		if(col[i] ^ -1) continue;
		memset(cnt, 0, sizeof(cnt)), tag = false, sum = 0;
		col[i] = 0, dfs(i);
		if(tag) ans += 1ll * (cnt[0] + cnt[1] + cnt[2]) * (cnt[0] + cnt[1] + cnt[2]);
		else if(!cnt[0] || !cnt[1] || !cnt[2]) ans += sum >> 1;
		else ans += 1ll * cnt[0] * cnt[1] + 1ll * cnt[1] * cnt[2] + 1ll * cnt[2] * cnt[0];
	}
	printf("%lld", ans);
}