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 s[205];
string fi[205], la[205];
bool dp[201][13][1 << 12];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    fi[i] = la[i] = s[i];
    for (int j = 1; j <= 12; j++) {
      if (j <= s[i].size()) {
        int val = 0;
        for (int k = 0; k < j; k++) {
          if (s[i][k] == '1') val += (1 << k);
        }
        dp[i][j][val] = 1;
        for (int k = j; k < s[i].size(); k++) {
          if (s[i][k - j] == '1') val--;
          val /= 2;
          if (s[i][k] == '1') val += (1 << (j - 1));
          dp[i][j][val] = 1;
        }
      }
    }
    if (fi[i].size() > 20) fi[i] = fi[i].substr(0, 20);
    if (la[i].size() > 20) la[i] = la[i].substr(la[i].size() - 20, 20);
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    n++;
    if (fi[a].size() < 20) {
      fi[n] = fi[a] + fi[b];
    } else {
      fi[n] = fi[a];
    }
    if (la[b].size() < 20) {
      la[n] = la[a] + la[b];
    } else {
      la[n] = la[b];
    }
    string mid = la[a] + fi[b];
    int ans = 0;
    for (int j = 1; j <= 12; j++) {
      if (j <= fi[n].size()) {
        int val = 0;
        for (int k = 0; k < j; k++) {
          if (fi[n][k] == '1') val += (1 << k);
        }
        dp[n][j][val] = 1;
        for (int k = j; k < fi[n].size(); k++) {
          if (fi[n][k - j] == '1') val--;
          val /= 2;
          if (fi[n][k] == '1') val += (1 << (j - 1));
          dp[n][j][val] = 1;
        }
      }
      if (j <= la[n].size()) {
        int val = 0;
        for (int k = 0; k < j; k++) {
          if (la[n][k] == '1') val += (1 << k);
        }
        dp[n][j][val] = 1;
        for (int k = j; k < la[n].size(); k++) {
          if (la[n][k - j] == '1') val--;
          val /= 2;
          if (la[n][k] == '1') val += (1 << (j - 1));
          dp[n][j][val] = 1;
        }
      }
      if (j <= mid.size()) {
        int val = 0;
        for (int k = 0; k < j; k++) {
          if (mid[k] == '1') val += (1 << k);
        }
        dp[n][j][val] = 1;
        for (int k = j; k < mid.size(); k++) {
          if (mid[k - j] == '1') val--;
          val /= 2;
          if (mid[k] == '1') val += (1 << (j - 1));
          dp[n][j][val] = 1;
        }
      }
      bool ok = 1;
      for (int k = 0; k < (1 << j); k++) {
        dp[n][j][k] |= dp[a][j][k];
        dp[n][j][k] |= dp[b][j][k];
        if (!dp[n][j][k]) ok = 0;
      }
      if (ok) ans = max(ans, j);
    }
    printf("%d\n", ans);
    if (fi[n].size() > 20) fi[n] = fi[n].substr(0, 20);
    if (la[n].size() > 20) la[n] = la[n].substr(la[n].size() - 20, 20);
  }
}
