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;
#define N 1116
#define M 1111116

int n,m,x[M],y[M],h[N][N],ans;

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();m=read();
	for (int i=1;i<=m;i++) x[i]=read(),y[i]=read();
	for (int i=1;i<=n;i++){
		h[i][i]=1;
		for (int j=m;j>=1;j--){
			if (!h[i][x[j]]&&!h[i][y[j]]) continue;
			if (h[i][x[j]]&&h[i][y[j]]) {fill(h[i]+1,h[i]+n+1,1);break;}
			h[i][x[j]]=h[i][y[j]]=1;
		}
	}
	for (int i=1;i<=n;i++)
		for (int j=i+1;j<=n;j++)
			for (int k=1;k<=n;k++)
				if (h[i][k]&&h[j][k])
				{ans++;break;};
	printf("%d\n",n*(n-1)/2-ans);
	return 0;
}