Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 2147483647;
const long double PI = acos(-1.0);
const int maxk = 10 + 1;
vector<vector<int> > calcK(string s) {
  vector<vector<int> > cd = vector<vector<int> >(maxk);
  for (auto k = (0); k != (maxk); ++k) {
    cd[k] = vector<int>(1 << maxk, 0);
    if (k > s.length()) continue;
    int b = 0;
    for (auto j = (0); j != (k); ++j) {
      b |= (s[j] == '1') * (1 << j);
    }
    cd[k][b] = 1;
    assert(b < (1 << k));
    if (k == 0) continue;
    for (auto j = (k); j != (s.length()); ++j) {
      b = (b >> 1);
      b |= (s[j] == '1') * (1 << (k - 1));
      assert(b < (1 << k));
      cd[k][b] = 1;
    }
  }
  return cd;
}
int cmk(vector<vector<int> > &cd) {
  for (int k = maxk - 1; k >= 0; k--) {
    int j = 0;
    for (; j < (1 << k); j++) {
      if (!cd[k][j]) break;
    }
    if (j < (1 << k)) continue;
    return k;
  }
  assert(false);
  return -1;
}
void run() {
  int n;
  cin >> n;
  vector<string> s(n);
  for (auto i = (0); i != (n); ++i) cin >> s[i];
  int m;
  cin >> m;
  int o = n + m;
  s.resize(o);
  vector<vector<int> > cds[o];
  int len[o];
  vector<string> first(o);
  vector<string> last(o);
  for (auto i = (0); i != (n); ++i) {
    cds[i] = calcK(s[i]);
    len[i] = min(20, (int)s[i].length());
    if (s[i].length() < 9)
      first[i] = last[i] = s[i];
    else {
      first[i] = last[i] = "";
      for (auto j = (0); j != (9); ++j) {
        first[i] += s[i][j];
        last[i] += s[i][s[i].length() - 9 + j];
      }
    }
  }
  for (auto i = (n); i != (o); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    cds[i] = vector<vector<int> >(maxk);
    if (len[a] < 20 && len[b] < 20) {
      s[i] = s[a] + s[b];
      len[i] = min(20, len[a] + len[b]);
      cds[i] = calcK(s[i]);
      if (s[i].length() < 9)
        first[i] = last[i] = s[i];
      else {
        first[i] = last[i] = "";
        for (auto j = (0); j != (9); ++j) {
          first[i] += s[i][j];
          last[i] += s[i][s[i].length() - 9 + j];
        }
      }
    } else {
      len[i] = 20;
      last[i] = last[a] + last[b];
      first[i] = first[a] + first[b];
      string fi = first[i], li = last[i];
      first[i] = last[i] = "";
      for (auto j = (0); j != (9); ++j) {
        first[i] += fi[j];
        last[i] += li[li.length() - 9 + j];
      }
      string mid = last[a] + first[b];
      vector<vector<int> > cd = calcK(mid);
      for (auto k = (0); k != (maxk); ++k) {
        cds[i][k] = vector<int>(1 << maxk, 0);
        for (auto j = (0); j != ((1 << k)); ++j)
          cds[i][k][j] = cd[k][j] | cds[a][k][j] | cds[b][k][j];
      }
    }
    cout << cmk(cds[i]) << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  run();
  return 0;
}