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;
long long n, i, j, k, x, y, z, m, q, h, r, timer, mx, l;
long long dp[205][14][4255];
string s[205], suf[205], pref[205];
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> s[i];
    long long sz = s[i].size();
    for (j = 0; j < sz; j++) {
      for (k = 1; k + j <= sz && k <= 12; k++) {
        long long cur = 0;
        for (l = j; l < j + k; l++) cur += (s[i][l] - '0') * (1 << (l - j));
        dp[i][k][cur] = 1;
      }
    }
    if (sz >= 12) {
      pref[i] = s[i].substr(0, 12);
      suf[i] = s[i].substr(sz - 12, 12);
    } else
      suf[i] = pref[i] = s[i];
  }
  cin >> m;
  long long num = n;
  for (i = 0; i < m; i++) {
    cin >> x >> y;
    string cur = suf[x];
    cur.append(pref[y]);
    num++;
    long long sz = cur.size();
    for (j = 0; j < sz; j++) {
      for (k = 1; k + j <= sz && k <= 12; k++) {
        long long tmp = 0;
        for (l = j; l < j + k; l++) tmp += (cur[l] - '0') * (1 << (l - j));
        dp[num][k][tmp] = 1;
      }
    }
    if (suf[x].size() + suf[y].size() <= 12)
      suf[num] = pref[num] = cur;
    else if (suf[x].size() >= 12 && suf[y].size() >= 12) {
      pref[num] = pref[x];
      suf[num] = suf[y];
    } else if (suf[x].size() >= 12) {
      pref[num] = pref[x];
      suf[num] = cur.substr(sz - 12, 12);
    } else if (pref[y].size() >= 12) {
      pref[num] = cur.substr(0, 12);
      suf[num] = suf[y];
    } else {
      pref[num] = cur.substr(0, 12);
      suf[num] = cur.substr(sz - 12, 12);
    }
    for (j = 12; j >= 1; j--) {
      for (l = 0; l < (1 << j); l++)
        dp[num][j][l] |= (dp[x][j][l] | dp[y][j][l]);
    }
    for (j = 12; j >= 1; j--) {
      long long cur = 0;
      for (l = 0; l < (1 << j); l++) cur += dp[num][j][l];
      if (cur == (1 << j)) {
        cout << j << endl;
        break;
      }
    }
    if (j == 0) cout << 0 << endl;
  }
  return 0;
}