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;
const double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int MR = 110;
struct Str {
  string p, s;
  set<string> S[MR];
  bool caly;
} t[2 * MR];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    string s;
    cin >> s;
    t[i].caly = 1;
    t[i].p = t[i].s = s;
    for (int dl = (1); dl <= (min((int)s.length(), 30)); ++dl)
      for (int b = 0; b < (s.length() - dl + 1); ++b)
        t[i].S[dl].insert(s.substr(b, dl));
  }
  int m;
  cin >> m;
  for (int i = 0; i < (m); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    string s = t[a].s + t[b].p;
    if (t[a].caly)
      t[i + n].p = s.substr(0, min(MR - 1, (int)s.length()));
    else
      t[i + n].p = t[a].p;
    if (t[b].caly) {
      if (s.length() < MR)
        t[i + n].s = s;
      else
        t[i + n].s = s.substr(s.length() - MR + 1);
    } else
      t[i + n].s = t[b].s;
    if (t[a].caly && t[b].caly && t[i + n].p == s && t[i + n].s == s)
      t[i + n].caly = 1;
    for (int j = (1); j < (31); ++j) {
      t[i + n].S[j].insert(t[a].S[j].begin(), t[a].S[j].end());
      t[i + n].S[j].insert(t[b].S[j].begin(), t[b].S[j].end());
    }
    for (int dl = (1); dl <= (min((int)s.length(), 30)); ++dl)
      for (int b = 0; b < (s.length() - dl + 1); ++b)
        t[i + n].S[dl].insert(s.substr(b, dl));
    int ile = 2, k = 0;
    for (int j = (1); j < (30); ++j) {
      if (t[i + n].S[j].size() == ile) k = j;
      ile <<= 1;
    }
    printf("%d\n", k);
  }
  return 0;
}
