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>
//this code is written by Itst
using namespace std;

int read(){
	int a = 0; char c = getchar();
	while(!isdigit(c)) c = getchar();
	while(isdigit(c)){
		a = a * 10 + c - 48; c = getchar();
	}
	return a;
}

bitset < 403 > dp[403];
int N , M , stp[100003][2];

int main(){
	N = read(); M = read();
	for(int i = 1 ; i <= M ; ++i){
		stp[i][0] = read(); stp[i][1] = read();
	}
	for(int i = 1 ; i <= N ; ++i){
		dp[i].set(i);
		for(int j = M ; j ; --j)
			if(dp[i][stp[j][0]] && dp[i][stp[j][1]]){
				dp[i].set(); break;
			}
			else
				if(dp[i][stp[j][0]] || dp[i][stp[j][1]]){
					dp[i].set(stp[j][0]); dp[i].set(stp[j][1]);
				}
	}
	int ans = 0;
	for(int i = 1 ; i <= N ; ++i)
		for(int j = i + 1 ; j <= N ; ++j)
			ans += (dp[i] & dp[j]).none();
	printf("%d\n" , ans);
	return 0;
}
