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 <cstring>
#include <algorithm>
#include <bitset>
using namespace std;
const int maxn = 405, maxm = 100005;
int N, M, a[maxm], b[maxm];
bitset<maxn> s[maxn];
int main() {
	scanf("%d %d", &N, &M);
	for(int i = 1; i <= M; ++i) 
		scanf("%d %d", a + i, b + i);
	for(int i = 1; i <= N; ++i) {
		bitset<maxn> &S = s[i];
		S[i] = 1;
		for(int j = M; j; --j) {
			if(S[a[j]] && S[b[j]]) {
				S[0] = 1; break;
			} 
			if(S[a[j]]) S[b[j]] = 1;
			if(S[b[j]]) S[a[j]] = 1;
		}
	}
	int ans = 0;
	for(int i = 1; i <= N; ++i) if(!s[i][0]) {
		for(int j = 1; j < i; ++j) if(!s[j][0]) {
			ans += not (s[i] & s[j]).any();
		}
	}
	printf("%d\n", ans);
	return 0;
}