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<cstdio>
#include<iostream>
#define re register
struct tur{
    int x,y;
}q[100001];
int c[501],n,m,ans,T,a[401][401];
inline int read() {
	char c=getchar();int x=0;while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-48,c=getchar();return x;
}
inline int check(int x,int y) {
	if(a[x][y]) return 0;
    ++T;c[x]=c[y]=T;
    for(re int i=m; i>=1; i--){
        if (c[q[i].x]==T&&c[q[i].y]==T) return 0;
        if (c[q[i].x]==T) c[q[i].y]=T;
        else if (c[q[i].y]==T) c[q[i].x]=T;
    }
    return 1;
}
int main(){
    n=read(),m=read();
    for(re int i=1; i<=m; i++)
        q[i].x=read(),q[i].y=read(),a[q[i].x][q[i].y]=a[q[i].y][q[i].x]=1;
    for(re int i=1; i<=n; i++)
        for(re int j=i+1; j<=n; j++)
            if(check(i,j)) ans++;
    printf("%d\n",ans);
    return 0;
}