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>
#define maxn 100010

using namespace std;

int n,m;

int x[maxn],y[maxn],ans;

bool f[510][510],S[510];

int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
		scanf("%d%d",&x[i],&y[i]);
	for (int i=1;i<=n;i++){
		f[i][i]=1;
		for (int j=m;j>=1;j--){
			int x1=f[i][x[j]],y1=f[i][y[j]];
			if (x1 && y1) {
				S[i]=1;
				break;
			}else if (x1) f[i][y[j]]=1;
			else if (y1) f[i][x[j]]=1;
		}
	}
	for (int i=1;i<=n;i++){
		if (S[i]==1) continue;
		for (int j=i+1;j<=n;j++){
			if (S[j]==1) continue;
			bool flag=1;
			for (int k=1;k<=n;k++)
				if (f[i][k] && f[j][k]){
					flag=0;
					break;
				}
			if (flag==1) ans++;
		}
	}
	printf("%d\n",ans);
	return 0;
}