// Original C++ solution (slower):
#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <vector>

#include <stack>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <cmath>

#include <cstdlib>

using namespace std;



#define M 1046527



typedef long long ll;



char H[M][20];



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;

}



ll getKey(char str[]) {

    ll 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[]) {

    ll key, i, h;

    key = getKey(str);

    for (int 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;

    }

}



int insert(char str[]) {

    ll key, i, h;

    key = getKey(str);

    for (int 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;

}



char cmd[10];

char str[10];

int n;



int main() {

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);

    scanf("%d", &n);

    while (n--) {

        scanf("%s %s", cmd, str);

        if (cmd[0] == 'i') {

            insert(str);

        } else {

            if (find(str)) {

                printf("yes\n");

            } else {

                printf("no\n");

            }

        }

    }

    return 0;

}

// Optimized solution to the same problem (faster):
/*

//alds1_4_c:Dictionary

//算法:开放地址法散列表

//Time: 2018/1/8 星期一

*/

#include<stdio.h>

#include<string.h>



const int M=1000003;

const int L=14;



char H[M][L];



//对于每个字符返回的定义值

int getChar(char ch){

    if(ch=='A') return 1;

    if(ch=='C') return 2;

    if(ch=='D') return 3;

    if(ch=='T') return 4;

    return 0;

}

//对于字符串返回的初始散列值

long long getKey(char str[]){

    long long len=strlen(str),sum=0,p=1;

    for(int i=0;i<len;++i){

        sum+=p*getChar(str[i]);

        //每次获取定义值后p*5,相当于转换成五进制,不会冲突

        p*=5;

    }

    return sum;

}



//开放式散列值计算式: h(k,i)=(h1(k)+i*h2(k))%M

int h1(int key){

    return key%M;

}

//为了保证不会递归冲突(即往下算结果始终相同),必须使h2(key)与M互素

//TLE最好的情况就是改这个函数= =

//目前可以AC的: 1+(key%(M-1))

//(1+key)%(M-1)

int h2(int key){

    return (1+key)%(M-1);

}



//查找

//-1表示找到

//h表示找到第一个可插入点

int find(char str[]){

    long long key=getKey(str),i,h;

    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 h;

    }

    return 0;

}



//插入

void insert(char str[]){

    int key=find(str);

    if(key!=-1) strcpy(H[key],str);

}



int main(){

    for(int i=0;i<M;++i) H[M][0]='\0';

    char str[L],com[L];

    int n;

    scanf("%d",&n);

    for(int i=0;i<n;++i){

        scanf("%s %s",com,str);



        if(com[0]=='i'){

            insert(str);

        }else{

            if(find(str)==-1)

                printf("yes\n");

            else

                printf("no\n");

        }

    }



    return 0;

}




// Fastest known solution for this problem: