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;
template <typename T>
T sqr(const T &first) {
  return first * first;
}
template <typename T>
T my_abs(const T &first) {
  return first < 0 ? -first : first;
}
const int INF = (int)1e9;
const long long INF64 = (long long)4e18;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
const int N = 205;
string s[N];
int n;
bool read() {
  cin >> n;
  for (int i = 0; i < int(n); ++i) cin >> s[i];
  return true;
}
const int M = 14;
pair<string, string> p[N];
bitset<1 << M> msk[N][M];
bool split[N];
void merge(int a, int b) {
  if (!split[a])
    p[n].first = p[a].first + p[b].first;
  else
    p[n].first = p[a].first;
  if (!split[b])
    p[n].second = p[a].second + p[b].first;
  else
    p[n].second = p[b].second;
  split[n] = split[a] | split[b];
  if (int(p[n].first.size()) > M) p[n].first.resize(M), split[n] = true;
  if (int(p[n].second.size()) > M) {
    reverse(p[n].second.begin(), p[n].second.end());
    p[n].second.resize(M);
    reverse(p[n].second.begin(), p[n].second.end());
    split[n] = true;
  }
  for (int i = 0; i < int(M); ++i) msk[n][i] = msk[a][i] | msk[b][i];
  string mid = p[a].second + p[b].first;
  for (int len = 0; len < int(M); ++len) {
    for (int pos = 0; pos < int(int(mid.size()) - len); ++pos) {
      string cc = "";
      int cmask = 0;
      for (int k = 0; k < int(len + 1); ++k) {
        cc += mid[pos + k];
        cmask = cmask * 2 + (mid[pos + k] - '0');
      }
      msk[n][len][cmask] = true;
    }
  }
  int cans = 0;
  for (int len = 0; len < int(M); ++len) {
    if (msk[n][len].count() == (2 << len)) {
      cans = len + 1;
    }
  }
  cout << cans << endl;
  n++;
}
void solve() {
  for (int i = 0; i < int(n); ++i) {
    p[i].first = p[i].second = s[i];
    for (int j = 0; j < int(M); ++j) {
      for (int pos = 0; pos < int(int(s[i].size()) - j); ++pos) {
        int cmask = 0;
        string cc;
        for (int k = 0; k < int(j + 1); ++k) {
          cmask = cmask * 2 + (s[i][pos + k] - '0');
          cc += s[i][pos + k];
        }
        msk[i][j][cmask] = true;
      }
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < int(m); ++i) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    merge(a, b);
  }
}
int main() {
  srand((unsigned int)time(NULL));
  cout << setprecision(15) << fixed;
  read();
  solve();
  return 0;
}
