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 <iostream>
#include <fstream>
#include <bits/stdc++.h> 
#include <string>

using namespace std;

typedef double db;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef priority_queue<int> pqi;
typedef priority_queue<int, vi, greater<int>> mpqi;

#define IOS std::ios::sync_with_stdio(false); cin.tie(0);
#define F0R(i,a) for(int i = 0; i < a; i++)
#define R0F(i,a) for(int i = a - 1; i >= 0; i--)
#define S0RT(a,n) sort(a, a + n)
#define pb push_back
#define mp make_pair

const int MOD = 1000000007;
const ll INF = 1000000000000000000;
const ld PI = 4*atan((ld)1);

/* NOTES TO SELF:
    Int overflow? Use all ll if in doubt.
    Edge cases? (n = 1, n = 0)
*/

int n;
int p[200002];

int main(){
    IOS;
    cin >> n;
    F0R(i,n){
        cin >> p[i];
    }
    int total = 0;
    int currMin = 99999999;
    F0R(i,n){
        if(p[i] < currMin){
            total++;
            currMin = p[i];
        }
    }
    cout << total;
    return 0;
}