#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 32761;
long long S[maxn];//存放的是S1,S2,到SK的和,S[5]表示了S1到S4的和,当数字变化到K的时候,一共有多少个字数了
int border[9] = {0,1,10,100,1000,10000,100000,1000000,10000000};
int num[9] = {0,9,90,900,9000,90000,900000,9000000,90000000};//num[1]存放1-9的位数,num[2]存放1-99的位数,以此类推,刚开始初始话成这样,是为了后面的计算
int getdigit(int n) {
int count = 0;
while(n > 0) {
count++;
n = n / 10;
}
return count;
}
void init() {
for(int i = 2; i < 9; i++)//位数的变化,num[2]存放的是1-99的位数,因为十位数要占两位,所以要用90个十位数*2,再加上前面的9个个位数,就是1-99的位数了,以此类推,num里面的下标表示的位数
num[i] = num[i-1] + num[i] * i;
S[0] = 0;
int digit;
for(int i = 1; i < maxn; i++) {
digit = getdigit(i);
S[i] = (i - border[digit] + 1 ) * digit + num[digit-1];
S[i] = S[i] + S[i-1];
}
}
int main() {
int n;
int N;
scanf("%d",&n);
init();
while(n--) {
scanf("%d",&N);
long long *p = lower_bound(S,S+maxn,N);
if(*p == N)
printf("%d\n",(p-S) % 10);
else {
int len = N - *(p-1);
bool flag = false;
int i;
for(i = 0; i < 9; i++) {
if(num[i] == len) {
flag = true;
break;
}
else if(num[i] > len)
break;
}
if(flag){
printf("9\n");
continue;
}
int last = len - num[i-1];
int k;
int count = i;
for(k = border[i]; count < last; count += i,k++);;
int time = count - last;
while(time > 0) {
k = k / 10;
time--;
}
printf("%d\n",k % 10);
}
}
return 0;
}