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;
void print(vector<int> a) {
  for (int x : a) printf("%d\n", x);
}
const int K = 16;
const int N = 1 << (K + 1);
vector<char> check(string s) {
  int n = s.length();
  vector<char> res(N);
  for (int i = 0; i < n; i++) {
    int chash = 1;
    for (int j = i; j < n && j - i + 1 <= K; j++) {
      chash = 2 * chash + (s[j] - '0');
      res[chash] = 1;
    }
  }
  return res;
}
vector<char> comb(vector<char> a, vector<char> b) {
  for (int i = 0; i < N; i++) {
    a[i] |= b[i];
  }
  return a;
}
string concatPref(string a, string b) {
  string res = a + b;
  if ((int)res.length() > K) res = res.substr(0, K);
  return res;
}
string concatSuf(string a, string b) {
  string res = a + b;
  if ((int)res.length() > K) res = res.substr((int)res.length() - K, K);
  return res;
}
int findLen(const vector<char> &f) {
  for (int len = 1; len <= K; len++) {
    bool ok = 1;
    for (int i = 0; i < (1 << len); i++) {
      int cmask = (1 << len) + i;
      ok &= f[cmask];
    }
    if (!ok) return len - 1;
  }
  return K + 1;
}
vector<int> solve(vector<string> s, vector<int> a, vector<int> b) {
  int n = s.size();
  vector<string> pref(n), suf(n);
  for (int i = 0; i < n; i++) {
    int len = min((int)s[i].length(), K);
    pref[i] = s[i].substr(0, len);
    suf[i] = s[i].substr((int)s[i].length() - len, len);
  }
  vector<vector<char> > hasMask(n, vector<char>(N));
  for (int i = 0; i < n; i++) {
    hasMask[i] = check(s[i]);
  }
  int m = a.size();
  vector<int> ans;
  for (int i = 0; i < m; i++) {
    vector<char> cres = comb(hasMask[a[i]], hasMask[b[i]]);
    cres = comb(cres, check(suf[a[i]] + pref[b[i]]));
    hasMask.push_back(cres);
    pref.push_back(concatPref(pref[a[i]], pref[b[i]]));
    suf.push_back(concatSuf(suf[a[i]], suf[b[i]]));
    ans.push_back(findLen(cres));
  }
  return ans;
}
void test() {
  vector<string> s(1);
  for (int i = 0; i < 100; i++) {
    int cnt = 0;
    int x = i;
    while (x) {
      cnt += x % 2;
      x /= 2;
    }
    s[0] += (char)('0' + cnt % 2);
  }
  vector<int> a, b;
  for (int i = 0; i < 100; i++) {
    a.push_back(i);
    b.push_back(i);
  }
  solve(s, a, b);
  exit(0);
}
int main() {
  int n;
  while (cin >> n) {
    vector<string> s(n);
    for (int i = 0; i < n; i++) {
      cin >> s[i];
    }
    int m;
    cin >> m;
    vector<int> a(m), b(m);
    for (int i = 0; i < m; i++) {
      cin >> a[i] >> b[i];
      a[i]--;
      b[i]--;
    }
    print(solve(s, a, b));
  }
  return 0;
}
