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>
typedef long long ll;
using namespace std;
#define repr(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define invrepr(i,a,b) for(int i=b-1;i>=a;i--)
#define invrep(i,n) invrepr(i,0,n)
#define repitr(itr,a) for(auto itr=a.begin();itr!=a.end();++itr)
const int MOD=1e9+7;



int main() {
    ios_base::sync_with_stdio(false); 

    int n;
    cin >> n;
    int p,mp;
    int ans=0;
    rep(i,n) {
        cin >> p;
        if (i==0) mp=p;
        if (p<=mp) ++ans;
        mp=min(mp,p);
    } 
    cout << ans << endl; 

    return 0;
}