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 dua[30], ans[205];
set<int> ls[205][20];
string suf[205], pre[205];
string getSuf(string x) {
  string ret = "";
  for (int i = (int)(x.length()) - 1; i >= 0; i--) {
    if (ret.length() == 20) break;
    ret = x[i] + ret;
  }
  return ret;
}
string getPre(string x) {
  string ret = "";
  for (int i = 0; i < (int)x.length(); i++) {
    if (i == 20) break;
    ret += x[i];
  }
  return ret;
}
void solve(string s, int idx, int fr) {
  int mx = s.length();
  mx = min(mx, 10);
  for (int i = fr; i <= mx; i++) {
    int r = i, l = 0;
    int now = 0;
    for (int j = 0; j < r; j++) {
      now *= 2;
      now += s[j] - '0';
    }
    ls[idx][i].insert(now);
    while (r < (int)s.length()) {
      now -= dua[i - 1] * (s[l] - '0');
      now *= 2;
      now += s[r] - '0';
      l++;
      r++;
      ls[idx][i].insert(now);
    }
  }
}
int main() {
  dua[0] = 1;
  for (int i = 1; i < 30; i++) dua[i] = dua[i - 1] * 2;
  int n;
  cin >> n;
  vector<string> ar;
  memset(ans, 0, sizeof ans);
  for (int i = 0; i < n; i++) {
    string a;
    cin >> a;
    ar.push_back(a);
    solve(a, i, 1);
    for (int j = 10; j >= 1; j--) {
      if (dua[j] == (int)ls[i][j].size()) {
        ans[i] = j;
        break;
      }
    }
    suf[i] = getSuf(a);
    pre[i] = getPre(a);
  }
  int m;
  cin >> m;
  for (int i = n; i < m + n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    int mx = max(ans[a], ans[b]);
    string nx = suf[a] + pre[b];
    if (ar[a].length() + ar[b].length() <= 300) {
      string temp = ar[a] + ar[b];
      suf[i] = getSuf(temp);
      pre[i] = getPre(temp);
      ar.push_back(temp);
    } else {
      suf[i] = suf[b];
      pre[i] = pre[a];
      if (ar[a].length() > ar[b].length())
        ar.push_back(ar[a]);
      else
        ar.push_back(ar[b]);
    }
    int sz = i;
    for (int j = mx + 1; j <= 10; j++) {
      set<int> &tmp = ls[sz][j];
      for (set<int>::iterator it = ls[a][j].begin(); it != ls[a][j].end();
           it++) {
        tmp.insert(*it);
      }
      for (set<int>::iterator it = ls[b][j].begin(); it != ls[b][j].end();
           it++) {
        tmp.insert(*it);
      }
    }
    solve(nx, sz, 1);
    ans[sz] = mx;
    for (int j = 10; j > mx; j--) {
      if (dua[j] == ls[sz][j].size()) {
        ans[sz] = j;
        break;
      }
    }
    cout << ans[sz] << endl;
  }
}
