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;

int read(){
	int a = 0; char c = getchar(); while(!isdigit(c)) c = getchar();
	while(isdigit(c)){a = a * 10 + c - 48; c = getchar();} return a;
}

const int _ = 5e5 + 7;
struct Edge{int end , upEd , w;}Ed[_ << 1];
int head[_] , cntEd , N , M , col[_] , num[3] , cnt;
void addEd(int a , int b , int c){Ed[++cntEd] = (Edge){b , head[a] , c}; head[a] = cntEd;}

bool dfs(int x , int c){
	++num[col[x] = c]; bool flg = 1;
	for(int i = head[x] ; i ; ++cnt , i = Ed[i].upEd)
		if(~col[Ed[i].end]) flg &= (c + Ed[i].w - col[Ed[i].end]) % 3 == 0;
		else flg &= dfs(Ed[i].end , (c + Ed[i].w) % 3);
	return flg;
}

int main(){
	N = read(); M = read(); memset(col , -1 , sizeof(col));
	for(int i = 1 ; i <= M ; ++i){int p = read() , q = read(); addEd(p , q , 1); addEd(q , p , 2);}
	long long sum = 0;
	for(int i = 1 ; i <= N ; ++i)
		if(col[i] == -1){
			num[0] = num[1] = num[2] = cnt = 0; bool flg = dfs(i , 0);
			if(!flg) sum += 1ll * (num[0] + num[1] + num[2]) * (num[0] + num[1] + num[2]);
			else if(!num[1] || !num[2]) sum += cnt >> 1;
			else sum += 1ll * num[0] * num[1] + 1ll * num[0] * num[2] + 1ll * num[1] * num[2];
		}
	cout << sum; return 0;
}
