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>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)

using namespace std;

const int N = 410, M = 1e5 + 10;

int n, m;
bool G[N][N], bad[N];
int X[M], Y[M];

int main(){
	scanf("%d%d", &n, &m);
	For(i, 1, m) scanf("%d%d", &X[i], &Y[i]);
	int ans = 0;

	For(i, 1, n){
		G[i][i] = true;
		Forr(j, m, 1){
			if(G[i][X[j]]){
				if(G[i][Y[j]]){
					bad[i] = true;
					break;
				}
				G[i][Y[j]] = true;
			}
			else if(G[i][Y[j]]) G[i][X[j]] = true;
		}

		if(!bad[i]) For(j, 1, i - 1) if(!bad[j]){
			bool flag = true;
			For(k, 1, n) if(G[i][k] && G[j][k]){
				flag = false;
				break;
			}
			ans += flag;
		}
	}

	printf("%d\n", ans);
	return 0;
}