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 B bitset<500>
using namespace std;
int n,m;
int x[200000],y[200000];
bool die[500];
B a[500];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
        scanf("%d%d",&x[i],&y[i]);
    for(int i=1;i<=n;i++)
    {
        a[i][i]=1;
        for(int j=m;j;j--)
        if(a[i][x[j]] && a[i][y[j]])
        {
            die[i]=1;
            break;
        }
        else
        if(a[i][x[j]])
            a[i][y[j]]=1;
        else
        if(a[i][y[j]])
            a[i][x[j]]=1;
    }
    int ans=0;
    for(int i=1;i<n;i++)
    if(!die[i])
        for(int j=i+1;j<=n;j++)
        if(!die[j] && (a[i]&a[j]).none())
            ++ans;
    printf("%d\n",ans);
    return 0;
}