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>
using namespace std;

int n,m; bitset<400> S[400];
int x[200005], y[200005];
bool no[200005];

int main() {
	scanf("%d%d", &n, &m);
	for (int i=0; i<n; i++) S[i][i] = 1;
	for (int i=0; i<m; i++) scanf("%d%d", &x[i], &y[i]), --x[i], --y[i];
	for (int i=m-1; i>=0; i--) {
		for (int j=0; j<n; j++) {
			if (S[j][x[i]] && S[j][y[i]]) no[j] = 1;
			else if (S[j][x[i]]) S[j][y[i]] = 1;
			else if (S[j][y[i]]) S[j][x[i]] = 1;
		}
	}
	int ans = 0;
	for (int i=0; i<n-1; i++)
		for (int j=i+1; j<n; j++)
			if (!no[i] && !no[j] && !(S[i] & S[j]).count()) {
				++ans;
			}
	printf("%d\n", ans);
	return 0;
}
 