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 <algorithm>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)

using namespace std;

const int N0 = 100010;
const int N = 410;

int x[N0], y[N0];
bool g[N][N], v[N];

int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	Rep(i, m) scanf("%d%d", &x[i], &y[i]);
	Rep(i, n) {
		g[i][i] = true;
		for (int j = m; j; j --) {
			if (!g[i][x[j]] && !g[i][y[j]]) continue;
			if (g[i][x[j]] && g[i][y[j]]) v[i] = true;
			else g[i][x[j]] = g[i][y[j]] = true;
		}
	}
	
	int ans = 0;
	
	Rep(i, n) if (!v[i]) 
		Rep(j, i - 1) if (!v[j]) {
			bool flag = false;
			Rep(k, n) if (g[i][k] && g[j][k]){ flag = true; break;}
			if (!flag) ans ++;
		}
	
	printf("%d\n", ans);
	
	return 0;
}