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 <vector>
#include <algorithm>

using namespace std;

#define all(v) v.begin(), v.end()
#define F first
#define S second
#define FOR(i, n) for(int i = 0; i < (int) n; i++)

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;

struct FT{
	int n;
	vi v;
	FT(int n){
		this->n = n;
		v.assign(n+1, 0);
	}
	int operator[](int i){
		int ans = 0;
		for(; i>0; i-=(i&(-i))) ans+=v[i];
		return ans; 
	}
	void add(int i, int x){
		for(; i <= n; i+=(i&(-i))) v[i]+=x;
	}
};

int main(){
	ios::sync_with_stdio(false); cin.tie(0);

	int n; cin >> n;
	FT f(n);
	int ans = 0;
	FOR(i, n){
		int x; cin >> x;
		if(f[x] == 0)
			ans++;
		f.add(x, 1);
	} 

	cout << ans << '\n';

	return 0;
}