Problem:
Score : 300 points

Problem Statement

Given is a permutation P_1, \ldots, P_N of 1, \ldots, N.
Find the number of integers i (1 \leq i \leq N) that satisfy the following condition:

For any integer j (1 \leq j \leq i), P_i \leq P_j.

Constraints

1 \leq N \leq 2 \times 10^5

P_1, \ldots, P_N is a permutation of 1, \ldots, N.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 ... P_N

Output

Print the number of integers i that satisfy the condition.

Sample Input 1

5
4 2 5 1 3

Sample Output 1

3

i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, P_i > P_j holds for j = 1.

Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.

Sample Input 2

4
4 3 2 1

Sample Output 2

4

All integers i (1 \leq i \leq N) satisfy the condition.

Sample Input 3

6
1 2 3 4 5 6

Sample Output 3

1

Only i=1 satisfies the condition.

Sample Input 4

8
5 7 4 2 6 8 1 3

Sample Output 4

4

Sample Input 5

1
1

Sample Output 5

1

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		int ans = 0;
		int be = Integer.MAX_VALUE;
		for(; n > 0; n--) {
			int ch = sc.nextInt();
			if(ch < be) {
				ans++;
				be = ch;
			}
			
		}
		pw.print(ans);
		pw.close();
	}
	
	static class FastScanner {
		private BufferedReader reader = null;
	    private StringTokenizer tokenizer = null;
	    public FastScanner(InputStream in) {
	        reader = new BufferedReader(new InputStreamReader(in));
	        tokenizer = null;
	    }

	    public String next() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }

	    public String nextLine() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                return reader.readLine();
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }

	        return tokenizer.nextToken("\n");
	    }

	    public long nextLong() {
	        return Long.parseLong(next());
	    }

	    public int nextInt() {
	        return Integer.parseInt(next());
	    }

	    public double nextDouble() {
	         return Double.parseDouble(next());
	     }

	    public int[] nextIntArray(int n) {
	        int[] a = new int[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextInt();
	        return a;
	    }

	    public long[] nextLongArray(int n) {
	        long[] a = new long[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextLong();
	        return a;
	    }
	}
}