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>

using namespace std;

#define N 600
#define M 120000

int n,m,a[M],b[M],ans;
bool f[N][N];

int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++)
        scanf("%d%d",&a[i],&b[i]);
    for (int i=1;i<=n;i++){
        f[i][i]=1;
        for (int j=m;j;j--){
            if (f[i][a[j]] && f[i][b[j]]){f[i][0]=1; break;}
            if (f[i][a[j]] || f[i][b[j]]) f[i][a[j]]=f[i][b[j]]=1;
        }
    }
    for (int i=1;i<=n;i++)
        for (int j=1;j<i;j++)
            if (!f[i][0] && !f[j][0]){
                bool fl=0;
                for (int k=1;k<=n;k++) fl|=(f[i][k]&f[j][k]);
                if (!fl) ans++;
            }
    printf("%d\n",ans);

    return 0;
}
