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 N = 202;
struct Item {
  string pre, suf;
  set<int> st[22];
  void print() { cout << pre << " " << suf << endl; }
  void init() {
    string s;
    cin >> s;
    int n = s.length();
    for (int i = 0; i < n; i++) {
      int mask = 0, len = 0;
      for (int j = i; j < n; j++) {
        mask = mask << 1 | (s[j] - '0');
        len++;
        if (len <= 20) st[len].insert(mask);
      }
    }
    for (int i = 0; i < min(n, 20); i++) pre = pre + s[i];
    for (int i = n - 1; i >= max(n - 20, 0); i--) suf = s[i] + suf;
  }
  int getAns() {
    int ans = 0;
    for (int i = 1;; i++) {
      if (i == 21)
        while (1)
          ;
      if (st[i].size() != (1 << i)) break;
      ans = i;
    }
    return ans;
  }
  Item operator+(const Item& o) const {
    Item ans;
    for (int i = 0; i < 22; i++) ans.st[i].clear();
    ans.pre = ans.suf = "";
    for (int i = 0; i < 22; i++) {
      for (auto x : st[i]) ans.st[i].insert(x);
      for (auto x : o.st[i]) ans.st[i].insert(x);
    }
    ans.pre = pre;
    ans.suf = o.suf;
    int l1 = 20 - pre.length();
    for (int i = 0; i < l1 && o.pre[i]; i++) ans.pre += o.pre[i];
    int l2 = 20 - o.suf.length();
    for (int i = suf.length() - 1; i >= 0; i--) {
      l2--;
      if (l2 < 0) break;
      ans.suf = suf[i] + ans.suf;
    }
    for (int i = suf.length() - 1; i >= 0; i--) {
      for (int j = 0; o.pre[j]; j++) {
        int mask = 0;
        for (int x = i; suf[x]; x++) mask = mask << 1 | (suf[x] - '0');
        for (int x = 0; x <= j; x++) mask = mask << 1 | (o.pre[x] - '0');
        if (suf.length() - i + j + 1 <= 20)
          ans.st[suf.length() - i + j + 1].insert(mask);
      }
    }
    return ans;
  }
} it[N];
int n, m;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) it[i].init();
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    it[i] = it[x] + it[y];
    cout << it[i].getAns() << endl;
  }
}