CodeForces #4C. Registration system

C. Registration system

time limit per test

5 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle.

Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name(name1, name2, ...), among these numbers the least i is found so that namei does not yet exist in the database.

Input

The first line contains number n (1 ≤ n ≤ 105). The following n lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters.

Output

Print n lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.

Examples

input

4
abacaba
acaba
abacaba
acab

output

OK
OK
abacaba1
OK

input

6
first
first
second
second
third
third

output

OK
first1
OK
second1
OK
third1

题目大意：

给出一些字符串，对每个字符串进行查询，若没出现过返回OK，若出现过就生成新字符串,格式为原字符串+数，数为这个字符串第几次重复出现。

解题思路：

字典树，对于每个字符串的插入次数进行计数。

#include <cstdio>
#include <cstring>
using namespace std;

struct node
{
    int p;
    int num[26];
}node[500010];

char s[50];
int cnt, len;

int index(int strp, int p)
{
    if (strp == len){
        node[p].p++;
        return node[p].p;
    }
    if (node[p].num[s[strp] - 'a'] == 0)
        node[p].num[s[strp] - 'a'] = cnt++;
    return index(strp + 1, node[p].num[s[strp] - 'a']);
}

int main()
{
    int n, temp;
    cnt = 1;
    memset(node, 0, sizeof(node));
    scanf("%d", &n);
    for (int i = 0; i < n; i++){
        scanf("%s", s);
        len = strlen(s);
        temp = index(0, 0);
        if (temp == 1)
            printf("OK\n");
        else{
            printf("%s%d\n", s, temp - 1);
        }
    }
    return 0;
}