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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <bitset>
using namespace std;
const int N=100010;
int a[N],b[N];
bitset <410> w[410];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int main() {
    int n,m,ans=0;
    cin>>n>>m;
    for(int i=1;i<=m;i++) a[i]=gi(),b[i]=gi();
    for(int i=1;i<=n;i++) {
	w[i][i]=1;
	for(int j=m;j;j--) {
	    int x=a[j],y=b[j];
	    if(w[i][x]&&w[i][y]) {w[i].set();break;}
	    if(w[i][x]) w[i].set(y);
	    else if(w[i][y]) w[i].set(x);
	}
    }
    for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++) ans+=((w[i]&w[j])==0);
    cout<<ans;
    return 0;
}