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;

template <typename T=int>
struct Fenwick {
    int n;
    vector<T> v;
    
    Fenwick(int size = 100005) : n(size), v(n+1, 0) {}
    inline void add(int p, T val) {
        for (; p <= n; p += (p&-p)) { v[p] += val; }
    }
    inline T query(int p) {
        T tmp = 0;
        for (; p > 0; p -= (p&-p)) { tmp += v[p]; }
        return tmp;
    }
    inline T query(int l, int r) {
        return query(r) - query(l-1);
    }
};

void solve() {
    int n; cin >> n;
    Fenwick<int> f(n);
    int cnt = 0;
    for (int i = 0; i < n; i++) {
        int x; cin >> x;
        if (f.query(x, n) == i) cnt++;
        f.add(x, 1);
    }
    cout << cnt;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}