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;

inline int gi()
{
	char c = getchar();
	while(c < '0' || c > '9') c = getchar();
	int sum = 0;
	while('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

const int maxn = 405, maxm = 100005;

int n, m;
int x[maxm], y[maxm];
bitset<maxn> dlt[maxn];
bool use[maxn], vis[maxn];

int main()
{
	n = gi(); m = gi();
	for (int i = 1; i <= m; ++i) x[i] = gi(), y[i] = gi();

	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) vis[j] = 0;
		dlt[i][i] = 1;
		for (int j = m; j >= 1; --j) {
			int a = dlt[i][x[j]], b = dlt[i][y[j]];
			if (a && b) {use[i] = 1; break;}
			else if (a) dlt[i][y[j]] = 1;
			else if (b) dlt[i][x[j]] = 1;
		}
	}
		
	int ans = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j)
			if (!use[i] && !use[j] && !(dlt[i] & dlt[j]).count() && !dlt[i][j]) ++ans;
	printf("%d\n", ans);
	
	return 0;
}

