Problem:
Search III

Your task is to write a program of a simple dictionary which implements the following instructions:

insert str: insert a string str in to the dictionary

find str: if the distionary contains str, then print 'yes', otherwise print 'no'

Input

In the first line n, the number of instructions is given. In the following n lines, n instructions are given in the above mentioned format.

Output

Print yes or no for each find instruction in a line.

Constraints

A string consists of 'A', 'C', 'G', or 'T'

1 ≤ length of a string ≤ 12

n ≤ 1000000

Sample Input 1

5
insert A
insert T
insert C
find G
find A

Sample Output 1

no
yes

Sample Input 2

13
insert AAA
insert AAC
insert AGA
insert AGG
insert TTT
find AAA
find CCC
find CCC
insert CCC
find CCC
insert T
find TTT
find T

Sample Output 2

yes
no
no
yes
yes
yes

Notes

Template in C

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

#define M 1046527
#define NIL (-1)
#define L 14

char H[M][L];

int getChar(char ch)
{
    if (ch == 'A') return 1;
    else if (ch == 'C') return 2;
    else if (ch == 'G') return 3;
    else if (ch == 'T') return 4;
    else return 0;
}

long long getKey(char str[]) {
    long long sum = 0, p = 1, i;
    for (int i = 0; i < strlen(str); i++) {
        sum += p*getChar(str[i]);
        p *= 5;
    }
    return sum;
}

int h1(int key) 
{
    return key % M;
}

int h2(int key)
{
    return 1 + (key % (M - 1));
}

int find(char str[])
{
    long long key, i, h;
    key = getKey(str);
    for (i = 0; ; i++) {
        h = (h1(key) + i * h2(key)) % M;
        if (strcmp(H[h], str) == 0) {
            return 1;
        } else if (strlen(H[h]) == 0) {
            return 0;
        }
    }
    return 0;
}

int insert(char str[])
{
    long long key, i, h;
    key = getKey(str);
    for (i = 0; ; i++) {
        h = (h1(key) + i * h2(key)) % M;
        if (strcmp(H[h], str) == 0) {
            return 1;
        } else if (strlen(H[h]) == 0) {
            strcpy(H[h], str);
            return 0;
        }
    }
    return 0;
}


int main(int argc, char const *argv[])
{
    int i, n, h;
    char str[L], com[9];
    for (int i = 0; i < M; i++) {
        H[i][0] = '\0';
    }
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%s %s", com, str);
        
        if (com[0] == 'i') {
            insert(str);
        } else {
            if (find(str)) {
                printf("yes\n");
            } else {
                printf("no\n");
            }
        }
    }

    return 0;
}