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>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

const int N = 405, M = 1e5 + 10;
int n, m, a[M], b[M], mark[N], ans;
bitset<N> s[N];

int main() {
    cin >> n >> m;
    rep(i, 1, m) scanf("%d%d", &a[i], &b[i]);
    rep(i, 1, n) {
        s[i][i] = 1;
        for (int j = m; j; --j) {
            if (s[i][a[j]] && s[i][b[j]]) { mark[i] = 1; break; }
            if (!s[i][a[j]] && !s[i][b[j]]) continue;
            s[i][a[j]] = s[i][b[j]] = 1;
        }
    }
    rep(i, 1, n)
        rep(j, i + 1, n)
            ans += (!mark[i] && !mark[j] && (s[i] & s[j]).none());
    printf("%d\n", ans);
    return 0;
}
