Problem:
Score : 300 points

Problem Statement

There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N. The elevation of Obs. i is H_i.
There are also M roads, each connecting two different observatories. Road j connects Obs. A_j and Obs. B_j.

Obs. i is said to be good when its elevation is higher than those of all observatories that can be reached from Obs. i using just one road.
Note that Obs. i is also good when no observatory can be reached from Obs. i using just one road.

How many good observatories are there?

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq H_i \leq 10^9

1 \leq A_i,B_i \leq N

A_i \neq B_i

Multiple roads may connect the same pair of observatories.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
H_1 H_2 ... H_N
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Print the number of good observatories.

Sample Input 1

4 3
1 2 3 4
1 3
2 3
2 4

Sample Output 1

2

From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.

From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.

From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.

From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.

Thus, the good observatories are Obs. 3 and 4, so there are two good observatories.

Sample Input 2

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6

Sample Output 2

3

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define llong long long
#define pb(a) push_back(a)
#define INF (long long)1 << 61
#define MOD 1000000007
using namespace std;

typedef pair<int, int> P;
typedef pair<llong, llong> LP;
typedef pair<int, P> PP;
typedef pair<llong, LP> LPP;
typedef vector<llong> LV;

int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

vector<pair<int,vector<int>>> H;

int reach(int n){
    H[n].first=0;
    for(auto e : H[n].second){
        // if(H[e].first>0){
        //     reach(e);
        // }
        // cout << "reach:" << e << endl;
        H[e].first = 0;
    }

    return 0;
}

int main() {
	int n,m; cin >> n >> m;

    REP(i,n){
        int h;cin >> h;
        vector<int> a;
        H.push_back(make_pair(h,a));
    }

    REP(i,m){
        int a,b;
        cin >> a >> b;
        H[a-1].second.push_back(b-1);
        H[b-1].second.push_back(a-1);
    }

    sort(H.begin(),H.end());

    int ans = 0;    
    REPR(i,n-1){
        if(H[i].first > 0){
            int isTop = 1;
            for(auto e : H[i].second){
                if(H[e].first >= H[i].first){
                    isTop = 0;break;
                }
            }
            if(isTop) ans++;
        }
        reach(i);
    }
    
    cout << ans << endl;

    return 0;
}