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;
typedef long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second

int n, m;
int qa[100000];
int qb[100000];
bitset<400> c[400], die;

int main()
{
	read(n, m);
	for(int i=0;i<m;++i)read(qa[i],qb[i]);
	for(int i=0;i<n;++i){
		auto&chk=c[i];
		chk.reset(); chk[i]=1;
		for(int j=m-1;0<=j;--j){
			int a=qa[j],b=qb[j];--a;--b;
			if(chk[a]&&chk[b]){die[i]=1;break;}
			else if(chk[a]) chk[b]=1;
			else if(chk[b]) chk[a]=1;
		}
	}
	int ans=0;
	for(int i=0;i<n;++i)if(!die[i])
	for(int j=i+1;j<n;++j)if(!die[j]) if((c[i]&c[j]).none())++ans;
	printf("%d\n", ans);
	return 0;
}
