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;

const int MAXN = 505, MAXM = 1e5 + 10;

int n, m;
long long ans;
int a[MAXM], b[MAXM];
bool jh[MAXN][MAXN], p[MAXN];

int main() {
	cin >> n >> m;
	for(int i = 1; i <= m; i++)
		cin >> a[i] >> b[i];
	
  	for(int i = 1; i <= n; i++) {
		p[i] = 1;
		jh[i][i] = 1;
		for(int j = m; j >= 1; j--) {
			int x = a[j], y = b[j];
			if(!jh[i][x] && !jh[i][y])
				continue;
			if(jh[i][x] && jh[i][y]) {
				p[i]=0;
				break;
			}
			jh[i][x] = jh[i][y] = 1;
		}
	}
	for(int i = 1; i <= n - 1; i++)
		for(int j = i + 1; j <= n; j++) {
			if(p[i] && p[j]) {
				int key = 0;
				for(int k = 1; k <= n; k++)
					if(jh[i][k] && jh[j][k]) {
						key  =1;
						break;
					}
				if(!key)
					ans++;
			}
		}
	cout << ans << '\n';
  	return 0;
}
