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 <iostream>
#include <fstream>
#include <bitset>
using namespace std;
const int nmax=405;
const int muchii=100005;
int x[muchii],y[muchii];
bitset<nmax> bs[nmax];
bool bad[nmax];
int n,m,i,j,pairs;
bool solve(bitset<nmax> &b)
{
    for(int i=m;i>=1;i--)
    {
        if(b[x[i]])
        {
            if(b[y[i]]) return 1;
            b[y[i]]=1;
        }
        if(b[y[i]])
        {
            b[x[i]]=1;
        }
    }
    return 0;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin>>n>>m;
    for(i=1;i<=m;i++)
    {
        cin>>x[i]>>y[i];
    }
    for(int x=1;x<=n;x++)
    {
        bs[x][x]=1;
        bad[x]=solve(bs[x]);
    }
    for(i=1;i<=n;i++)
        for(j=i+1;j<=n;j++)
           if((!bad[i])&&(!bad[j])&&(bs[i]&bs[j]).count()==0)
                 pairs++;
    cout<<pairs;
    return 0;
}