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>

using namespace std;

int a[400][400];
int b[400];

int main() {
    int n, m, ans = 0, i, j;
    
    scanf("%d %d", &n, &m);
    
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            a[i][j] = 1;
        }
    }
    
    for (i = 0; i < m; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        x--;
        y--;
        
        if (a[x][y] == 0 || b[x] == 1 && b[y] == 1) b[x] = b[y] = 1;
        
        a[x][y] = a[y][x] = 0;
        
        for (j = 0; j < n; j++) {
            if (a[x][j] == 0 || a[y][j] == 0 || b[x] == 1 || b[y] == 1) {
                a[x][j] = a[j][x] = a[y][j] = a[j][y] = 0;
            }
        }
    }
    
    for (i = 0; i < n; i++) {
        for (j = i + 1; j < n; j++) {
            ans += a[i][j];
        }
    }
    
    printf("%d\n", ans);
    
    return 0;
}
