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: Wrong Answer)

#include <iostream>
using namespace std;
#include <math.h>

int main(void){
    int a=0 ,i=0,j=0;
    bool flag = true;
    scanf("%d",&a);
    printf("%d\n",a);
    int *c;        /* 割り振られた領域のアドレスを格納するためのint型のポインタipを宣言 */
        c = (int *)malloc(a * sizeof(int));  /* メモリ領域の確保 */

    for(i=0;i<a;i++){
        scanf("%d",&c[i]);
        printf("%d ",c[i]);
    }
    
    int count = 0;
    flag = true;
    for(i=0;i<a;i++){
        flag = true;
        //i番目の数より前の数字にそれより大きいものがないか確認する
        for(j=0;j<=i;j++){
            if(c[i]>c[j]){
                flag = false;
            }
        }
        if(flag){
            count += 1;
        }
    }
    printf("\n%d",count);
    free(c);
}