Problem: 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.

Examples

Input

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


Output

2


Input

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


Output

3

Solution (in JAVA):

Solution:

import java.util.Scanner;

/**
 * Created with IntelliJ IDEA.
 * User: ngupta
 * Date: 5/5/2020 AD
 * Time: 00:33
 */
class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        int[] height = new int[N + 1];
        int[] ma = new int[N + 1];
        for (int i = 1; i <= N; i++) {
            height[i] = sc.nextInt();
        }

        for (int i = 0; i < M; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            ma[x] = Math.max(ma[x], height[y]);
            ma[y] = Math.max(ma[y], height[x]);
        }

        int count = 0;
        for (int i = 1; i < ma.length; i++) {
            if (height[i] > ma[i])
                count++;
        }

        System.out.println(count);
    }
}