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 pair<int, int> pii;

int N, M;
pii P[402];
vector<pii> adj[402];
int vis[402];
int ans;

bool dfs(int u, int fe) {
    if(vis[u]) return false;
    vis[u] = 1;

    for(int i = 0; i < adj[u].size(); i++) {
        int v = adj[u][i].first;
        int e = adj[u][i].second;
        if(e >= fe) continue;
        if(vis[v]) return false;
        if(!dfs(v, e)) return false;
    }
    return true;
}

int main() {
    scanf("%d %d", &N, &M);

    for(int i = 0; i < M; i++) {
        int u, v; scanf("%d %d", &u, &v);
        u--; v--;

        adj[u].push_back({v, i});
        adj[v].push_back({u, i});
    }
    for(int i = 0; i < N; i++) {
        for(int j = i + 1; j < N; j++) {
            memset(vis, 0, sizeof(vis));
            if(!dfs(i, 1e9)) continue;
            if(!dfs(j, 1e9)) continue;
            ans++;
        }
    }
    printf("%d", ans);
}
