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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

int main() {
	int n,c=1,d,flag=0,p;
	
	
	
	
	
	 
        scanf("%d",&n);
        int a[n];
        for(int i=0;i<n;i++){
           scanf("%d",&a[i]);
        }
        
         for(int i=0;i<n;i++){
	     if(a[i]==1){
	         p=i;
	         break;
	     }
	 }
        if(p==0){
            printf("1");
        }
        else{
         for(int i=1;i<=p;i++){
            
            flag=0;
            for(int j=0;j<i;j++){
                if(a[j]<a[i]){
                    flag=1;
                    break;
                }
                }
            if(flag==0){
                c++;
            }
            }
       printf("%d",c);
        }
	return 0;
}