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(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 i,n,h;
  char str[L],com[9];
  for(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;
}