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<algorithm>
using namespace std;
int gi(){
	int x=0,w=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')w=0,ch=getchar();
	while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
const int N=405;
int n,m,x[N*N],y[N*N],f[N][N],g[N],ans;
int work(int *dp){
	for(int i=m;i;--i){
		if(dp[x[i]]&&dp[y[i]])return 1;
		if(dp[x[i]]||dp[y[i]])dp[x[i]]=dp[y[i]]=1;
	}
	return 0;
}
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)f[i][i]=1,g[i]=work(f[i]);
	for(int i=1;i<=n;++i)
		if(!g[i])
			for(int j=i+1;j<=n;++j)
				if(!g[j]){
					int fg=1;
					for(int k=1;k<=n;++k)fg&=!(f[i][k]&f[j][k]);
					ans+=fg;
				}
	printf("%d\n",ans);return 0;
}
