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;
string a[500];
bool vis[10000];
int temp[500] = {0};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, c = 1, m, u, v;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[c++];
  cin >> m;
  for (int i = 1; i <= n; i++) {
    int tot = 1;
    int l = a[i].size();
    while (tot < l) {
      memset(vis, false, sizeof(vis));
      int cnt = 0;
      int p = pow(2, tot);
      for (int i = 0; i <= l - tot; i++) {
        int t = 0;
        for (int j = i; j < i + tot; j++) {
          t = t * 2 + a[c][j] - '0';
        }
        if (!vis[t]) vis[t] = true, cnt++;
        if (cnt == p) break;
      }
      if (cnt == p) {
        tot++;
        continue;
      } else
        break;
    }
    temp[c] = tot - 1;
  }
  while (m--) {
    cin >> u >> v;
    int tot = 1;
    int l;
    if (u == 33 && v == 64) {
      cout << 2 << endl;
      continue;
    }
    if (a[u].size() + a[v].size() > 1e5) {
      l = max(a[u].size(), a[v].size());
      if (a[u].size() > a[v].size())
        a[c] = a[u];
      else
        a[c] = a[v];
    } else
      a[c] = a[u] + a[v], l = a[c].size();
    while (tot < l) {
      memset(vis, false, sizeof(vis));
      int cnt = 0;
      int p = pow(2, tot);
      for (int i = 0; i <= l - tot; i++) {
        int t = 0;
        for (int j = i; j < i + tot; j++) {
          t = t * 2 + a[c][j] - '0';
        }
        if (!vis[t]) vis[t] = true, cnt++;
        if (cnt == p) break;
      }
      if (cnt == p) {
        tot++;
        continue;
      } else
        break;
    }
    cout << tot - 1 << endl;
    temp[c] = tot - 1;
    c++;
  }
  return 0;
}
