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<bitset>
using namespace std;

const int N = 400;
const int M = 1e5;
bitset<N+3> f[N+3],tmp;
int a[M+2],b[M+2];
int n,m;

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] = true;
        for(int j=m; j>=1; j--)
        {
            bool x = f[i][a[j]],y = f[i][b[j]];
            if(x && y) {f[i].set(0); break;}
            if((!x) && y) {f[i].set(a[j]);}
            else if(x && (!y)) {f[i].set(b[j]);}
        }
    }
    int ans = 0;
    for(int i=1; i<=n; i++)
    {
        for(int j=i+1; j<=n; j++)
        {
            if(f[i][0] || f[j][0]) continue;
            tmp = f[i]&f[j];
            if(!(tmp.any())) ans++;
		}
	}
	printf("%d\n",ans);
    return 0;
}