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;
const int N = 400, Q = 1e5 + 5;
bitset <N> s[N];
int a[Q], b[Q];
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, q; cin >> n >> q;
    for (int i = 1; i <= q; i++) cin >> a[i] >> b[i], a[i]--, b[i]--;
    for (int i = 0; i < n; i++) {
        s[i][i] = 1;
        for (int j = q; j > 0; j--) {
            if (s[i][a[j]] == 1 && s[i][b[j]] == 1) {s[i].set(); continue ;}
            s[i][a[j]] = s[i][b[j]] = (s[i][a[j]] | s[i][b[j]]);
        }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) ans += ((s[i]&s[j]).count() == 0);
    cout << ans << "\n";

    return 0;
}
