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;
#define pii pair<int,int>
#define ff first
#define ss second
#define sfi(a) scanf("%d",&a)
#define sfi2(a,b) scanf("%d %d",&a,&b)
#define sfi3(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfll(a) scanf("%lld",&a)
#define all(a) a.begin(),a.end()
#define keepu(a) a.erase(unique(all(a)),a.end())
#define inf 1e7
#define m_p make_pair
#define min3(a,b,c) min(a,min(b,c))
#define pb push_back
#define mem(x,val) memset(x,val,sizeof(x))
#define read freopen("output.txt","r",stdin)
#define write freopen("input.txt","w",stdout)
#define FAST  ios_base::sync_with_stdio(0);cout.tie(0)
#define sqr(x) ((x)*(x))

typedef long long ll;
const ll mod = 1e9+7;
const int sz = 500;


int main(){

    int n;
    cin >> n;
    int arr[n];
    for(int i=0;i<n;i++) cin >> arr[i];
    int mn = arr[0];

    int cnt = 1;

    for(int i=1;i<n;i++){
        if(arr[i]<=mn){
            cnt++;
        }
        mn = min(mn,arr[i]);
    }

    cout << cnt << endl;

    return 0;
}