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<cstdlib>
#include<cstring>
#include<algorithm>
#include<bitset>
#define MAXN 410
int x[100010], y[100010], n, m, ans;
std::bitset<MAXN> z[MAXN];
int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1;i <= m;i++)scanf("%d%d", &x[i], &y[i]);
    for(int i = 1;i <= n;i++){
        z[i].set(i);
        for(int j = m;j;j--){
            if(z[i][x[j]] && z[i][y[j]]){
                z[i].set(0);
                break;
            }else if(z[i][x[j]])z[i].set(y[j]);
            else if(z[i][y[j]])z[i].set(x[j]);
        }
    }
    for(int i = 1;i <= n;i++){
        for(int j = i + 1;j <= n;j++){
            if(!z[i][0] && !z[j][0] && (z[i] & z[j]).none())ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}