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 <algorithm>
#include <cctype>
#include <bitset>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 4e2+5, M = 1e5+5;
bitset<N> s[N];
int n, m, u[M], v[M], d[N], ans;
int main()
{
    n = read(), m = read();
    for(int i = 1; i <= m; ++i) u[i] = read(), v[i] = read();
    for(int i = 1; i <= n; ++i)
    {
        s[i][i] = 1;
        for(int j = m; j; --j)
            if(s[i][u[j]]&&s[i][v[j]]) d[i] = 1, j = 1;
            else if(s[i][u[j]]) s[i][v[j]] = 1;
            else if(s[i][v[j]]) s[i][u[j]] = 1;
    }
    for(int i = 1; i <= n; ++i)
    {
        if(d[i]) continue;
        for(int j = i+1; j <= n; ++j)
            if(!d[j]&&(s[i]&s[j]).none()) ++ans;
    }
    printf("%d\n", ans);
    return 0;
}