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<cstdio>
#include<bitset>
using namespace std;
const int MAXN = 405, MAXM = 100005;
int n, m, ans, x[MAXM], y[MAXM], vis[MAXN];
bitset<MAXN> S[MAXN];

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) {
		S[i].set(i);
		for(int j = m; j >= 1; --j)
		if(S[i][x[j]] && S[i][y[j]]) vis[i] = 1;
		else if(S[i][x[j]]) S[i].set(y[j]);
		else if(S[i][y[j]]) S[i].set(x[j]);
	}
	
	for(int i = 1; i <= n; ++i) if(!vis[i])
	for(int j = i + 1; j <= n; ++j) if(!vis[j])
		if((S[i]&S[j]).none()) ++ans;
	printf("%d\n", ans);
	return 0;
}