题目：

Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.

He has names of n people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.

Formally, for a name si in the i-th line, output "YES" (without quotes) if there exists an index j such that si = sj and j < i, otherwise, output "NO" (without quotes).

Input

First line of input contains an integer n (1 ≤ n ≤ 100) — the number of names in the list.

Next n lines each contain a string si, consisting of lowercase English letters. The length of each string is between 1 and 100.

Output

Output n lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.

You can print each letter in any case (upper or lower).

Examples

input

6
tom
lucius
ginny
harry
ginny
harry

output

NO
NO
NO
NO
YES
YES

input

3
a
a
a

output

NO
YES
YES

Note

In test case 1, for i = 5 there exists j = 3 such that si = sj and j < i, which means that answer for i = 5 is "YES".

题意：

  给出n次询问，问之前的询问中是否出现个这个字符串。

思路：

  map一下就可以。

实现：

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr); int _;
    map<string, int> mp; string str;
    while(mp.clear(), cin >> _)
        while(_--) {
            cin >> str;
            puts(mp[str]++ ? "YES" : "NO");
        }
    return 0;
}