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;
const int N=450; 
const int M=1e5+100;
bool f[N][N];
int n,m,u[M],v[M];
bool chck[N];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&u[i],&v[i]);
    }
    for(int i=1;i<=n;i++){
        f[i][i]=1;
        for(int j=m;j>=1;j--){
            bool x=f[i][u[j]],y=f[i][v[j]];
            if(x&&y){
			chck[i]=1;
			break;
			}
            else if(x) f[i][v[j]]=1;
            else if(y) f[i][u[j]]=1;
        }
    }
    int ans=0;
    for(int i=1;i<n;i++){
        if(chck[i]) continue;
        for(int j=i+1;j<=n;j++){
            if(chck[j]) continue;
            bool fl=1;
            for(int k=1;k<=n;k++){ 
                if(f[i][k]&&f[j][k]) fl=0;
            } 
            ans+=fl;
        }
    }
    printf("%d",ans);
    return 0;
}