You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.

Input

The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.

The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.

Output

Print m lines, each should contain one integer — the answer to the question after the corresponding operation.

Example

Input

5
01
10
101
11111
0
3
1 2
6 5
4 4


Output

1
2
0

Note

On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.

On the second operation the string "01100" is created. Now all strings of length k = 2 are present.

On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
#include <bits/stdc++.h>
using namespace std;
int n, ex[205][13][1 << 13], cnt[205][13], len[205];
char buf[205];
int rd[205];
int pref[205][13], suff[205][13];
int m;
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    scanf("%s", buf);
    int m = strlen(buf);
    len[i] = m;
    for (int j = (0); j < (m); j++) rd[j] = buf[j] - '0';
    for (int j = (0); j < (m); j++) {
      int cur = 0;
      for (int ln = (1); ln < (13); ln++) {
        if (j + ln > m) continue;
        cur = 2 * cur + rd[j + ln - 1];
        ex[i][ln][cur] = 1;
        if (j == 0) pref[i][ln] = cur;
        if (j + ln == m) suff[i][ln] = cur;
      }
    }
  }
  scanf("%d", &m);
  for (int i = (0); i < (m); i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    int at = n + i;
    for (int l = (1); l < (13); l++)
      for (int j = (0); j < (1 << l); j++)
        ex[at][l][j] = (ex[a][l][j] | ex[b][l][j]);
    for (int sl = (1); sl < (min(13, len[a] + 1)); sl++)
      for (int pl = (1); pl < (min(13, len[b] + 1)); pl++) {
        if (sl + pl >= 13) continue;
        int val = suff[a][sl] * (1 << pl) + pref[b][pl];
        ex[at][pl + sl][val] = 1;
      }
    for (int l = (1); l < (min(13, len[a] + 1)); l++) pref[at][l] = pref[a][l];
    for (int l = (len[a] + 1); l < (min(13, len[a] + len[b] + 1)); l++) {
      pref[at][l] = pref[a][len[a]] * (1 << (l - len[a])) + pref[b][l - len[a]];
    }
    for (int l = (1); l < (min(13, len[b] + 1)); l++) suff[at][l] = suff[b][l];
    for (int l = (len[b] + 1); l < (min(13, len[a] + len[b] + 1)); l++) {
      suff[at][l] = suff[a][l - len[b]] * (1 << (len[b])) + suff[b][len[b]];
    }
    len[at] = min(13, len[a] + len[b]);
    for (int j = (1); j < (13); j++) {
      int g = 1;
      for (int k = (0); k < (1 << j); k++)
        if (!ex[at][j][k]) g = 0;
      if (!g) {
        printf("%d\n", j - 1);
        break;
      }
    }
  }
  return 0;
}
