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;
using bs = bitset<(1 << 15)>;
void insert(bs &b, int length, int value) {
  if (length > 14) return;
  b[(1 << length) - 2 + value] = true;
}
vector<bs> goals;
int findMax(bs &b) {
  for (int i = 15 - 1; i > 0; i--) {
    if ((b & goals[i]) == goals[i]) return i;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  goals.resize(15);
  for (int i = 1; i < 15; i++) {
    goals[i].reset();
    for (int j = 0; j < (1 << i); j++) {
      insert(goals[i], i, j);
    }
  }
  int n;
  cin >> n;
  vector<bs> v(201);
  vector<string> starts(201), ends(201);
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    starts[i] = s;
    ends[i] = s;
    v[i].reset();
    int sz = s.size();
    for (int a = 0; a < sz; a++) {
      int value = 0;
      for (int b = a; b < sz; b++) {
        value = 2 * value + (s[b] - '0');
        insert(v[i], b + 1 - a, value);
      }
    }
  }
  int Q;
  cin >> Q;
  for (int q = 1; q <= Q; q++) {
    int x, y;
    cin >> x >> y;
    v[n + q] = v[x] | v[y];
    if (starts[x].size() < 14) {
      string sss = starts[x] + starts[y];
      starts[n + q] = sss;
    } else {
      starts[n + q] = starts[x];
    }
    if (ends[y].size() < 14) {
      string sss = ends[x] + ends[y];
      ends[n + q] = sss;
    } else {
      ends[n + q] = ends[y];
    }
    auto &s1 = ends[x];
    auto &s2 = starts[y];
    int sz1 = s1.size();
    int sz2 = s2.size();
    for (int a = max(0, sz1 - 15); a < sz1; a++) {
      int value = 0;
      int length = 0;
      for (int b = a; b < sz1 + min(15, sz2); b++) {
        length++;
        value = 2 * value + (b < sz1 ? s1[b] - '0' : s2[b - sz1] - '0');
        insert(v[n + q], length, value);
      }
    }
    cout << findMax(v[n + q]) << endl;
  }
}