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;
int gi(){
    char ch=getchar(); int x=0,q=0;
    while(ch<'0'||ch>'9') q=ch=='-'?1:q,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return q?-x:x;
}
const int N=401,M=1e5+50;
bitset<N>f[N];
int a[M],b[M],vis[N];
int main(){
    int n=gi(),m=gi(),x,y;
    for (int i=1; i<=m; ++i)
	a[i]=gi(),b[i]=gi();
    for (int i=1; i<=n; ++i)
	f[i].set(i);
    for (int i=m; i>=1; --i){
	x=a[i],y=b[i];
	for (int t=1; t<=n; ++t)
	    if (f[t][x]||f[t][y]){
		if (f[t][y]) swap(x,y);
		if (f[t][y]) vis[t]=1;
		else f[t].set(y);
	    }
    }
    int ans=0;
    for (int i=1; i<=n; ++i)
	if (!vis[i])
	    for (int j=i+1; j<=n; ++j)
		if (!vis[j]&&(f[i]&f[j]).count()==0)
		    ++ans;
    cout<<ans;
    return 0;
}
