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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 233;
int n, m;
struct str {
  static const int K = 10;
  static const int SIZE = 1 << K + 1;
  string pre, suf;
  bitset<SIZE> b;
  void adjust() {
    if (((int)(pre).size()) > K) pre = pre.substr(0, K);
    if (((int)(suf).size()) > K) suf = suf.substr(((int)(suf).size()) - K, K);
  }
} a[N];
str add(str a, str b) {
  str c;
  c.b = a.b | b.b;
  if (((int)(c.pre = a.pre).size()) < str::K) c.pre += b.pre;
  if (((int)(c.suf = b.suf).size()) < str::K) c.suf = a.suf + b.suf;
  c.adjust();
  string g = a.suf + b.pre;
  for (int i = 0, ed = ((int)(g).size()); i < ed; ++i) {
    int t = 0;
    for (int j = i, ed = ((int)(g).size()); j < ed; ++j) {
      t = t << 1 | (g[j] - '0');
      int tp = (1 << j - i + 1) | t;
      if (tp >= str::SIZE) break;
      c.b[tp] = 1;
    }
  }
  return c;
}
int solve(str a) {
  for (int i = str::K + 1 - 1, ed = 1; i >= ed; --i) {
    bool ok = 1;
    for (int j = 0, ed = 1 << i; j < ed; ++j) {
      if (!a.b[1 << i | j]) {
        ok = 0;
        break;
      }
    }
    if (ok) return i;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, ed = n; i < ed; ++i) {
    string s;
    cin >> s;
    a[i].pre = a[i].suf = s;
    for (int j = 0, ed = ((int)(s).size()); j < ed; ++j) {
      int t = 0;
      for (int k = j, ed = ((int)(s).size()); k < ed; ++k) {
        t = t << 1 | (s[k] - '0');
        int tp = (1 << k - j + 1) | t;
        if (tp >= str::SIZE) break;
        a[i].b[tp] = 1;
      }
    }
  }
  cin >> m;
  for (int i = 0, ed = m; i < ed; ++i) {
    int j, k;
    cin >> j >> k;
    a[n] = add(a[j - 1], a[k - 1]);
    cout << solve(a[n++]) << endl;
  }
  return 0;
}
