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 n;
vector<string> v(200 + 200, "");
vector<string> suf(200 + 200, "");
vector<string> pre(200 + 200, "");
bool has[202][8][129];
int deg[222];
int cg(string &s) {
  int ret = 0;
  for (auto i = 0; i < (int)s.size(); i++) {
    ret = ret << 1 | (s[i] - '0');
  }
  return ret;
}
bool check(int idx, int now) {
  for (int i = 0; i < 1 << now; i++) {
    if (!has[idx][now][i]) return 0;
  }
  return 1;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(has, 0, sizeof has);
    memset(deg, 0, sizeof deg);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      for (int len = 1; len <= 7; len++) {
        for (int st = 0; st + len - 1 < (int)v[i].size(); st++) {
          string tmp = v[i].substr(st, len);
          has[i][len][cg(tmp)] = 1;
        }
      }
      int tmplen = min((int)v[i].size(), 7);
      pre[i] = v[i].substr(0, tmplen);
      suf[i] = v[i].substr(v[i].size() - tmplen, tmplen);
      deg[i] = 0;
      while (deg[i] < 7 && check(i, deg[i] + 1)) deg[i]++;
    }
    int m;
    scanf("%d", &m);
    int cur = n;
    while (m--) {
      int l, r;
      scanf("%d%d", &l, &r);
      l--, r--;
      for (int len = 1; len <= 7; len++)
        for (int i = 0; i < 128; i++)
          if (has[l][len][i] || has[r][len][i]) has[cur][len][i] = 1;
      for (int len = 2; len <= 7; len++) {
        for (int x = 1; x <= (int)suf[l].size() && x < len; x++) {
          int y = len - x;
          if (y > (int)pre[r].size()) continue;
          auto tmpstr =
              suf[l].substr(suf[l].size() - x, x) + pre[r].substr(0, y);
          has[cur][len][cg(tmpstr)] = 1;
        }
      }
      deg[cur] = 0;
      while (deg[cur] < 7 && check(cur, deg[cur] + 1)) deg[cur]++;
      if (suf[l].size() < 7) {
        pre[cur] = suf[l] + pre[r];
        if (pre[cur].size() > 7) pre[cur] = pre[cur].substr(0, 7);
      } else
        pre[cur] = pre[l];
      if (pre[r].size() < 7) {
        int nedlen = min(7 - (int)pre[r].size(), (int)suf[l].size());
        suf[cur] = suf[l].substr(suf[l].size() - nedlen, nedlen) + pre[r];
      } else
        suf[cur] = suf[r];
      printf("%d\n", deg[cur]);
      cur++;
    }
  }
  return 0;
}
