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 INF = 0x3f3f3f3f;
const int MAXN = 300;
const int MAXK = 14;
int n, a, b, m;
string l[MAXN], r[MAXN];
set<int> hs[MAXN][MAXK];
int get_int(string p) {
  int ret = 0;
  for (int i = (0); i < (p.length()); i++) {
    ret = ret * 2 + (p[i] - '0');
  }
  return ret;
}
void get_hs(string &str, int p) {
  for (int k = (1); k < (MAXK); k++) {
    for (int i = 0; i + k <= str.length(); i++) {
      int t = get_int(str.substr(i, k));
      hs[p][k].insert(t);
    }
  }
}
int main() {
  cin >> n;
  for (int i = (1); i < (n + 1); i++) {
    cin >> l[i];
    get_hs(l[i], i);
    if (l[i].length() > MAXK)
      r[i] = l[i].substr(l[i].length() - MAXK, MAXK),
      l[i] = l[i].substr(0, MAXK);
    else
      r[i] = l[i];
  }
  cin >> m;
  for (int i = (1); i < (m + 1); i++) {
    cin >> a >> b;
    string s = r[a] + l[b];
    if (l[a].length() < MAXK) {
      if (s.length() >= MAXK)
        l[i + n] = s.substr(0, MAXK);
      else
        l[i + n] = s;
    } else
      l[i + n] = l[a];
    if (r[b].length() < MAXK) {
      if (s.length() >= MAXK)
        r[i + n] = s.substr(s.length() - MAXK, MAXK);
      else
        r[i + n] = s;
    } else
      r[i + n] = r[b];
    for (int j = (1); j < (MAXK); j++) {
      for (auto &x : hs[a][j]) {
        hs[i + n][j].insert(x);
      }
      for (auto &x : hs[b][j]) {
        hs[i + n][j].insert(x);
      }
    }
    get_hs(s, i + n);
    for (int j = (1); j < (MAXK); j++) {
      if (hs[i + n][j].size() < (1 << j)) {
        printf("%d\n", j - 1);
        break;
      }
    }
  }
  return 0;
}