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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using lint     = long long;
const lint inf = 1LL << 60;
const lint mod = 1000000007;

struct UnionFind {
    vector<int> rank, parent, size;

    // +1 for 1-indexed nodes
    UnionFind(int n) : rank(n + 1, 0), parent(n + 1), size(n + 1, 1) {
        iota(parent.begin(), parent.end(), 0); // parent is itself
    }
    int root(int x) {
        if (x != parent[x]) {
            parent[x] = root(parent[x]);
        }
        return parent[x];
    }
    bool isSame(int x, int y) { return root(x) == root(y); }
    bool unite(int x, int y) { return link(root(x), root(y)); }
    bool link(int x, int y) {
        if (x == y)
            return false;
        if (rank[x] > rank[y]) {
            parent[y] = x;
            size[x] += size[y];
        } else {
            parent[x] = y;
            size[y] += size[x];
            if (rank[x] == rank[y]) {
                rank[y]++;
            }
        }
        return true;
    }
    int getSize(int x) { return size[root(x)]; }
};

template <class T>
bool chmax(T &a, const T &b) {
    return (a < b) ? (a = b, 1) : 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    return (b < a) ? (a = b, 1) : 0;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    lint n, m;
    cin >> n >> m;
    vector<lint> h(n);
    for (int i = 0; i < n; ++i) {
        cin >> h[i];
    }
    vector<vector<int>> edges(n);
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        u--, v--;
        edges[u].emplace_back(v);
        edges[v].emplace_back(u);
    }
    int ret = 0;
    for (int i = 0; i < n; ++i) {
        bool ok = true;
        for (auto &v : edges[i]) {
            if (h[v] >= h[i]) {
                ok = false;
                break;
            }
        }
        if (ok)
            ret++;
    }
    cout << ret << "\n";
    return 0;
}