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;

int main(){
    int N, M;
    cin >> N >> M;
    vector<int> X(M), Y(M);
    for(int i=0; i<M; i++){
        cin >> X[i] >> Y[i];
        X[i]--; Y[i]--;
    }
    
    vector<bitset<400>> S(N);
    for(int i=0; i<N; i++){
        S[i][i] = 1;
        for(int j=M-1; j>=0; j--){
            int x = X[j], y = Y[j];
            if(S[i][x] && S[i][y]){
                S[i].set();
                break;
            }else if(S[i][x] || S[i][y]){
                S[i][x] = S[i][y] = 1;
            }
        }
    }

    int ans = 0;
    for(int i=0; i<N; i++) for(int j=0; j<i; j++) if((S[i] & S[j]).none()) ans++;
    cout << ans << endl;
    return 0;
}
