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
//2018-1-15
//miaomiao
//
#include <bits/stdc++.h>
using namespace std;

#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)
#define N (400+5)

bool dead[N], vis[N][N];

int main(){
	int n, m, x, y;

	scanf("%d%d", &n, &m);
	For(ti, 1, m){
		scanf("%d%d", &x, &y);

		if(vis[x][y]) dead[x] = dead[y] = true;
		if(dead[x]) dead[y] = true; if(dead[y]) dead[x] = true;
		vis[x][y] = vis[y][x] = true;
		
		For(i, 1, n) if(i != x && i != y){
			if(vis[i][x]) vis[i][y] = vis[y][i] = true;
			if(vis[i][y]) vis[i][x] = vis[x][i] = true;
		}
	}

	int ans = 0;
	For(i, 1, n) if(!dead[i]) For(j, i+1, n)
		if(!dead[j]) ans += !vis[i][j];
	printf("%d\n", ans);

	return 0;
}