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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x,long long int y) 
{ 
    long long int res = 1;     // Initialize result 
  // x=x;
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y & 1) 
            res = (res*x); 
   
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x);  // Change x to x^2 
    } 
    return res; 
}
long long int power(long long int x,long long int y,long long int m) 
{ 
    long long int res = 1;     // Initialize result 
   x=x%m;
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y & 1) 
            res = (res*x)%m; 
   
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x)%m;  // Change x to x^2 
    } 
    return res; 
}

long long int gcd(long long int a, long long int b){ 
    if (a == 0) 
        return b;  
    return gcd(b % a, a);  
 } 
 long long int lcm(long long int a, long long int b)  
 { 
     long long int temp;
    
    return (a*b)/gcd(a, b);  
 }
 
int main() {
        long long int t=0;
        string c;
        string s;
       long long int n,m,o=0,l,d=0,i,j,p=0,k,f;
       double pie=3.141592653589793238;
    long long int MOD=1000000007;
    cin>>n;
    long long int a[n],b[n];
    for(i=0;i<n;i++)
    {
        cin>>a[i];
        b[a[i]-1]=i;
    }
    //d=d+1;
    p=n;
    for(i=0;i<n;i++)
    {
        if(b[i]<p)
        {
            //cout<<i<<" ";
            d=d+1;
            p=b[i];
        }
    }
    cout<<d;
    //cout<<P;
 return 0;
}