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>
using namespace std;

int N, M;
int rel[410][410];

int main() {
    scanf("%d%d", &N, &M);
    for(int i=0; i<M; i++) {
        int x, y; scanf("%d%d", &x, &y);
        x--; y--;

        for(int z=0; z<N; z++) {
            if(rel[x][z]) rel[y][z] = true;
            if(rel[y][z]) rel[x][z] = true;
            if(rel[z][x]) rel[z][y] = true;
            if(rel[z][y]) rel[z][x] = true;
        }
        if(rel[y][y]) rel[x][x] = true;
        if(rel[x][x]) rel[y][y] = true;
        rel[x][y] = rel[y][x] = true;
    }

    int ans = 0;
    for(int i=0; i<N; i++) {
        for(int j=i+1; j<N; j++) {
            if(rel[i][i] || rel[j][j]) continue;
            if(!rel[i][j] && !rel[j][i]) {
                ans++;
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}