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 <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAXN = 410;
const int MAXM = 100010;

int n, m, a[MAXM], b[MAXM];

void input() {
	scanf( "%d%d", &n, &m );
	for( int i = 0; i < m; ++i )
		scanf( "%d%d", a+i, b+i );
}

int f[MAXN][MAXN] = {0}, ok[MAXN];
bool check( int a, int b ) {
	for( int i = 1; i <= n; ++i )
		if( f[a][i] && f[b][i] ) return false;
	return true;
}
void solve() {
	for( int i = 1; i <= n; ++i ) {
		ok[i] = 1, f[i][i] = 1;
		for( int j = m-1; j >= 0; --j ) {
			int x = a[j], y = b[j];
			if( f[i][x] && f[i][y] ) {
				ok[i] = 0; break;
			}
			if( !f[i][x] && !f[i][y] ) continue;
			f[i][x] = f[i][y] = 1;
		}
	}
	int ans = 0;
	for( int i = 1; i <= n; ++i )
		for( int j = i+1; j <= n; ++j )
			if( ok[i] && ok[j] && check(i,j) )
				++ans;
	printf( "%d\n", ans );
}

int main() {
	input(), solve();
	return 0;
}
