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: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;


#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep1(i, n) for (int i = 1; i < (n); i++)
#define rrep(i, n) for (int i = (n)-1; i >= 0; i--)
#define all(i) (i).begin(), (i).end()
#define rall(i) (i).begin(), (i).end(), greater<int>()

#define int long long

#define elif else if
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define fst first
#define sec second

#define mod 1000000007

template <typename T>
void print(T x)
{
	std::cout << x <<" " ;
}


template <typename T>
void println(T x)
{
	std::cout << x << "\n";
}

int RepeatSquaring(int n,int x) {
	if (x == 0) return 1;
	else if (x % 2 == 0) {
		int div = RepeatSquaring(n, x / 2);
		return div*div%mod;
	}
	return n*RepeatSquaring(n,x-1)%mod;
}

int c(int n, int a) {
	int x = 1;
	int y = 1;
	rep(i, a) {
		y *= a - i;
		x *= n - i;
		y %= mod;
		x %= mod;
	}
	return (x*RepeatSquaring(y,mod-2))%mod;
}


signed main(void) {
	int n;
	cin >> n;
	vector<int>p;
	p.resize(n);
	rep(i, n) {
		cin >> p[i];
	}
	int ans = 0;
	rep(i, n) {
		rep(j, i+1) {
			if (p[j] < p[i])break;
			if(i==j) ans++;
		}
	}
	print(ans);
}