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 <bits/stdc++.h>
#define oo 405
using namespace std;
int n, m, vis[oo], tot;
int mat[oo][oo];
int main ()
{
    scanf("%d%d", &n, &m);
    for(int i = 1, x, y; i <= m; ++ i)
    {
        scanf("%d%d", &x, &y);
        if(vis[x]) vis[y] = 1;
        else if(vis[y]) vis[x] = 1;
        else if(mat[x][y]) vis[x] = vis[y] = 1;
        else
        {
            mat[x][y] = mat[y][x] = 1;
            for(int j = 1; j <= n; ++ j)
            if(mat[x][j]) mat[y][j] = mat[j][y] = 1;
            for(int j = 1; j <= n; ++ j)
            if(mat[y][j]) mat[x][j] = mat[j][x] = 1;
        }
    }
    for(int i = 1; i <= n; ++ i)
    for(int j = i + 1; j <= n; ++ j)
    if(!vis[i] && !vis[j] && !mat[i][j]) ++ tot;
    printf("%d", tot); return 0;
}