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 R register
#define IN inline
#define W while
#define gc getchar()
#define ll long long
#define MX 405
#define SIZ 100500
template <class T>
IN void in(T &x)
{
	x = 0; R char c = gc;
	for (; !isdigit(c); c = gc);
	for (;  isdigit(c); c = gc)
	x = (x << 1) + (x << 3) + c - 48;
}
int n, m, ans;
int a[SIZ], b[SIZ];
std::bitset <MX> st[MX];
bool out[MX];
int main(void)
{
	in(n), in(m);
	for (R int i = 1; i <= m; ++i)
	in(a[i]), in(b[i]);
	for (R int i = 1; i <= n; ++i)
	{
		st[i][i] = true;
		for (R int j = m; j; --j)
		{
			if (st[i][a[j]] && st[i][b[j]])
			{out[i] = true; goto gg;}
			if ((!st[i][a[j]]) && (!st[i][b[j]])) continue;
			st[i][a[j]] = st[i][b[j]] = true;
		}
		gg: ;
	}
	for (R int i = 1; i < n; ++i)
	for (R int j = i + 1; j <= n; ++j)
	ans += (!out[i]) && (!out[j]) && (st[i] & st[j]).none();
	printf("%d", ans);
}
