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;

#define ll long long
#define pb push_back

const int maxn = 4e2 + 20;
const int maxm = 1e5 + 20;

bitset<maxn> adj[maxn] , tmp;

int x[maxm] , y[maxm];

bool has[maxn];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n , m;
	cin >> n >> m;

	for(int i = 0; i < m; i++)
	{
		cin >> x[i] >> y[i];
		x[i]-- , y[i]--;
	}

	memset(has , 1 , sizeof has);

	for(int i = 0; i < n; i++)
		for(int j = m - 1; j >= 0; j--)
		{
			adj[i][i] = 1;
			if(adj[i][x[j]] && adj[i][y[j]])
			{
				has[i] = 0;
				break;
			}

			adj[i][x[j]] = adj[i][y[j]] = adj[i][x[j]] | adj[i][y[j]];
		}

	int res = 0;

	for(int i = 0; i < n; i++)
		for(int j = i + 1; j < n; j++)
			if(has[i] && has[j])
			{
				tmp = adj[i]&adj[j];

				res += !tmp.any();
			}

	cout << res << endl;
}



