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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int main()
{
	int n;
	scanf("%d", &n);
	int arr[200001];
	int i;
	for (i = 0; i < n; i++)
	{
		scanf("%d", &arr[i]);
	}
	int sum = 1;
	int j;
	for (i = 1; i < n; i++)
	{
		if (arr[i] > arr[i - 1])
			continue;
		if (arr[i] == 1)
		{
			sum++;
			break;
		}
		int f = 0;
		for (j = i - 1; j >= 0; j--)
		{
			if (arr[i] > arr[j])
				f = 1;
		}
		if (f == 0)
			sum++;
	}
	printf("%d", sum);
	return 0;
}