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;
bool stu[505][505], c[505];
int n, m, a[100005],b[100005], ans;
int main(int argc, char **argv) {
    cin >> n >> m;
    for(int i=1;i<=m;i++)
        cin >> a[i] >> b[i];
    for(int i=1;i<=n;i++) {
        stu[i][i]=1;
        for(int j=m;j>=1;j--) {
            bool x=stu[i][a[j]],y=stu[i][b[j]];
            if(x!=0&&y!=0) {
                c[i]=1;
                break;
            }
            else if(x) stu[i][b[j]]=1;
            else if(y) stu[i][a[j]]=1;
        }
    }
    for(int i=1;i<n;i++) {
        if(c[i]) continue;
        for(int j=i+1;j<=n;j++) {
            if(c[j]) continue;
            bool flag=1;
            for(int k=1;k<=n;k++)
                if(stu[i][k]&&stu[j][k]) flag=0;
            ans+=flag;
        }
    }
    cout << ans << endl;
}