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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
//if(strcmp(s[i],"AC")==0)
//if(s[l]=='#')
/*小文字判定
if((moji>= 'a') && (moji<='z')){
    printf("a\n");
    return 0;
  }
  */

int W_Length(char *str){
	int length = 0;
	while(*str++ != '\0')
		length++;
	return length;
}

void swap(long long int *a, long long int *b){
  int temp;
  temp=*a;
  *a=*b;
  *b=temp;
}
//swap(&a, &b)

double deg(double theta){//deg to theta
return theta*(2*3.14159265359)/360;
}

long long int cmp( const void *p, const void *q ) {
    return *(long long int*)p - *(long long int*)q;
}//ちいさい順
//qsort(a,n,sizeof(long long int),cmp);


int cmp_s(const void * left, const void * right)
{
	char *left_char = (char *)left;
  char *right_char = (char *)right;
  return strcmp(left_char, right_char);
}
//qsort(s,n,sizeof(s[0]),cmp_s);
//char s[n][11]

long long int GCD(long long int a, long long int b){
  long long int L=a,S=b;
  while(1){
    if(S>L){
      L=S+L;
      S=L-S;
      L=L-S;
    }
    if(L%S==0){
      return S;
    }
    L=L%S;
  }
  
}

long long int LCM(long long int a, long long int b){
  return a*b/(GCD(a,b));
}

int main (void)
{
	long long int p[5005],b,c,n,i,min,k,a[210000],max,j=1,t[210000],count=0;
scanf("%lld",&n);
for(i=0;i<n;i++){
  scanf("%lld",&a[i]);
}  
  min=a[0];
  for(i=0;i<n;i++){
    if(a[i]<=min){
      min=a[i];
      count++;
    }
  }
  
  printf("%lld",count);
return 0;
}