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;
int n, m;
const int N = 500;
const int M = 200000;
int xi[M], yi[M];
int f[N][N], GG[N], ans;
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; ++ i)
        cin >> xi[i] >> yi[i];
    for (int i = 1; i <= n; ++ i)
    {
        f[i][i] = 1;
        for (int j = m; j >= 1; -- j)
        {
            if (f[i][xi[j]] + f[i][yi[j]] == 2)
                GG[i] = 1;
            if (f[i][xi[j]] + f[i][yi[j]] == 0)
                continue;
            f[i][xi[j]] = f[i][yi[j]] = 1;
        }
    }
    for (int i = 1; i <= n; ++ i) if (!GG[i])
        for (int j = i + 1; j <= n; ++ j) if (!GG[j])
        {
            int cnt = 1;
            for (int k = 1; k <= n; ++ k)
                if (f[i][k] && f[j][k])
                    cnt = 0;
            ans += cnt;
        }
    cout << ans << endl;
}
