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>

#define NR 410

bool un[NR][NR], del[NR];
int n, m;

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; ++i) {
		int x, y;
		scanf("%d%d", &x, &y);
		
		if (del[x] || del[y] || un[x][y] || un[y][x]) {
			if (!del[x] || !del[y]) {
				del[x] = del[y] = true;
				for (int k = 1; k <= n; ++k) un[k][x] = un[k][y] = true;
			}
		} else {
			for (int k = 1; k <= n; ++k) if (k != x && k != y) {
				if (un[k][x]) un[k][y] = true;
				if (un[x][k]) un[y][k] = true;
				if (un[k][y]) un[k][x] = true;
				if (un[y][k]) un[x][k] = true;
			}
			
			un[x][y] = true;
			un[y][x] = true;
		}
	}
	
	int cnt = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j < i; ++j)
			if (!un[i][j] && !un[j][i])
				++cnt;
				
	printf("%d\n", cnt);
	return 0;
}