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;
template <class T, class T2>
inline int chkmax(T &x, const T2 &y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class T2>
inline int chkmin(T &x, const T2 &y) {
  return x > y ? x = y, 1 : 0;
}
const int MAXN = (1 << 8);
const int MAXK = (1 << 18);
struct state {
  string pref, suff;
  bitset<MAXK> bs[20];
  state() {}
  state(string s) {
    pref = suff = s;
    while (pref.size() > 20) pref.pop_back();
    reverse(suff.begin(), suff.end());
    while (suff.size() > 20) suff.pop_back();
    reverse(suff.begin(), suff.end());
    bs[0][0] = 1;
    for (int i = 0; i < (int)s.size(); i++) {
      int mask = 0;
      for (int l = 1; l <= 18 && i - l + 1 >= 0; l++) {
        mask |= (s[i - l + 1] - '0') << (l - 1);
        bs[l][mask] = 1;
      }
    }
  }
  void push_back(const state &oth) {
    for (int i = 0; i < 20; i++) bs[i] |= oth.bs[i];
    string mid = suff + oth.pref;
    string nw_pref = pref, nw_suff = oth.suff;
    if (nw_pref.size() < 20) nw_pref = nw_pref + oth.pref;
    if (nw_suff.size() < 20) nw_suff = suff + nw_suff;
    while (nw_pref.size() > 20) nw_pref.pop_back();
    reverse(nw_suff.begin(), nw_suff.end());
    while (nw_suff.size() > 20) nw_suff.pop_back();
    reverse(nw_suff.begin(), nw_suff.end());
    pref = nw_pref;
    suff = nw_suff;
    state oth2 = state(mid);
    for (int i = 0; i < 20; i++) bs[i] |= oth2.bs[i];
  }
};
int n;
string s[MAXN];
state st[MAXN];
void read() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    st[i] = state(s[i]);
  }
}
void solve() {
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    st[++n] = st[x];
    st[n].push_back(st[y]);
    int ans = 0;
    while ((int)st[n].bs[ans].count() == (1 << ans)) ans++;
    cout << ans - 1 << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}