Problem:
Score : 300 points

Problem Statement

Given is a sequence of integers A_1, A_2, ..., A_N.
If its elements are pairwise distinct, print YES; otherwise, print NO.

Constraints

2 ≤ N ≤ 200000

1 ≤ A_i ≤ 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

If the elements of the sequence are pairwise distinct, print YES; otherwise, print NO.

Sample Input 1

5
2 6 1 4 5

Sample Output 1

YES

The elements are pairwise distinct.

Sample Input 2

6
4 1 3 1 6 2

Sample Output 2

NO

The second and fourth elements are identical.

Sample Input 3

2
10000000 10000000

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#define MAX 2000001

void quick(int *,int,int);

int main() {
    int k,n;
    static int a[MAX];
    scanf("%d", &n);
    for (k = 0; k < n; ++k) {
        scanf("%d", &a[k]);
    }
    quick(a,0,n-1);
    for (k = 0; k < n-1; ++k) {
        if(a[k]==a[k+1]){
            printf("NO");
            return 0;
        }
    }
    printf("YES");
    return 0;
}
void quick(int a[],int left, int right){
    int s,t,i,j;

    if(left<right){
        s=a[left];  //左端の項を軸にする
        i=left; j=right+1;  //軸より小さいグループと大きいグループに分ける
        while(1){
            while(a[++i]<s);
            while(a[--j]>s);
            if(i>=j) break;
            t=a[i];a[i]=a[j];a[j]=t;
        }
        a[left]=a[j];a[j]=s;    //軸を正しい位置に入れる

        quick(a,left,j-1);  //左部分列に対する再起呼び出し
        quick(a,j+1,right); //右部分列に対する再起呼び出し
    }
}