There are N turkeys. We number them from 1 through N.

M men will visit here one by one. The i-th man to visit will take the following action:

* If both turkeys x_i and y_i are alive: selects one of them with equal probability, then eats it.
* If either turkey x_i or y_i is alive (but not both): eats the alive one.
* If neither turkey x_i nor y_i is alive: does nothing.



Find the number of pairs (i,\ j) (1 ≤ i < j ≤ N) such that the following condition is held:

* The probability of both turkeys i and j being alive after all the men took actions, is greater than 0.

Constraints

* 2 ≤ N ≤ 400
* 1 ≤ M ≤ 10^5
* 1 ≤ x_i < y_i ≤ N

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the number of pairs (i,\ j) (1 ≤ i < j ≤ N) such that the condition is held.

Examples

Input

3 1
1 2


Output

2


Input

4 3
1 2
3 4
2 3


Output

1


Input

3 2
1 2
1 2


Output

0


Input

10 10
8 9
2 8
4 6
4 9
7 8
2 8
1 8
3 4
3 4
2 7


Output

5
#include <cstdio>
#include <iostream>
#include <algorithm>
const int N = 405;
int n, m, ans;
bool w[N][N];
inline int Get() {
	char ch;
	while ((ch = getchar()) < '0' || ch > '9');
	int Num = ch - '0';
	while ((ch = getchar()) >= '0' && ch <= '9')
		Num = (Num << 3) + (Num << 1) + ch - '0';
	return Num;
}
int main() {
	n = Get(), m = Get();
	for (int k = 1; k <= m; ++k) {
		int x = Get(), y = Get();
		if (w[x][y]) {
			for (int i = 1; i <= n; ++i) w[x][i] = w[i][x] = true;
			for (int i = 1; i <= n; ++i) w[y][i] = w[i][y] = true;
			continue;
		}
		
		w[x][y] = w[y][x] = true;
		for (int i = 1; i <= n; ++i)
			if (i != x && i != y) {
				if (w[i][x] && !w[i][y]) w[i][y] = w[y][i] = true;
				if (w[i][y] && !w[i][x]) w[i][x] = w[x][i] = true;
			}
		
	}
	
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j < i; ++j) ans += !w[i][j];
	
	printf("%d\n", ans);
}