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: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#include <math.h>

#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;



int main() {

    int n;
    cin >> n;
    vector<int> p(n);
    rep(i, n) cin >> p[i];
    int cnt = 0;
    int ans = 1;

    if(p[1]<=p[0])
     {
        ans += 1;
        sort(p.begin(), p.begin() + 1);
    }
    for (int i = 2; i < n;i++)
    {
        if (p[i]<=p[i-1])
        {
            p.erase(p.begin() + i);
            ans += 1;
            i--;
            n--;
        } 
    }

    cout << ans << endl;

    return 0;
}