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<cmath>
#include<bitset>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 100005;
bitset <405> a[405];
int n, m, i, j, k, ans;
int x[MAXN], y[MAXN], gg[MAXN];
int main()
{
	cin >> n >> m;
	for(i = 1; i <= m; i ++)
		scanf("%d%d", &x[i], &y[i]);
	for(i = 1; i <= n; i ++)
	{
		a[i][i] = 1;
		for(j = m; j >= 1; j --)
			if (a[i][x[j]] && a[i][y[j]]){gg[i] = 1; break;}
			else if (a[i][x[j]]) a[i][y[j]] = 1;
			else if (a[i][y[j]]) a[i][x[j]] = 1;
	}
	for(i = 1; i <= n; i ++)
		if (!gg[i])
			for(j = 1 + i; j <= n; j ++)
				if (!gg[j])
					if ((a[i] & a[j]).count() == 0)
						ans ++;
	cout << ans << endl;
}