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 N 405
#define M 100050
using namespace std;
int F[N][N],imp[N],a[M],b[M],n,m;
inline int rd() {int r;scanf("%d",&r);return r; }
int main() {
	n = rd(), m = rd();
	for (int _=1;_<=m;_++) a[_] = rd(), b[_] = rd();

	for (int x=1;x<=n;x++) {
		F[x][x] = 1;
		for (int _=m;_>=1;_--) {
			if (!F[x][a[_]] && !F[x][b[_]]) continue;
			if (F[x][ a[_] ] && F[x][ b[_] ]) {imp[x] = 1; break;}
			if (F[x][ a[_] ])
				F[x][ b[_] ] = 1;
			else
				F[x][ a[_] ] = 1;
		}
	}
	
	int ans = 0;
	for (int i=1;i<=n;i++)
		for (int j=i+1;j<=n;j++) if (!imp[i] && !imp[j]) {
			int flag = 1;
			for (int _=1;_<=n;_++) if (F[i][_] && F[j][_]) flag = 0;
			ans += flag;
		}
	cout << ans << endl;
	return 0;
}