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=410;
const int M=1e5+10;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,m,x[M],y[M],c[N];
bitset<N> s[N];
int main() {
    n=gi(),m=gi();
    for(int i=1;i<=m;i++) x[i]=gi(),y[i]=gi();
    for(int i=1;i<=n;i++) {
        c[i]=1;
        s[i][i]=1;
        for(int j=m;j;j--) {
            if(!s[i][x[j]]&&!s[i][y[j]]) continue;
            if(s[i][x[j]]&&s[i][y[j]]) { c[i]=0;break; }
            s[i][x[j]]=s[i][y[j]]=1;
        }
    }
    int ans=0;
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            if(c[i]&&c[j]&&!(s[i]&s[j]).count()) ++ans;
    printf("%d\n",ans);
    return 0;
}
