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;
const int N = 210;
string s[N];
string p;
int ans[N];
int nex[N];
int lenq;
void GetNext(string q, int len) {
  memset(nex, 0, sizeof(nex));
  nex[0] = -1;
  int k = -1;
  for (int i = 1; i < len; i++) {
    while (k != -1 && q[k + 1] != q[i]) k = nex[k];
    if (q[k + 1] == q[i]) k++;
    nex[i] = k;
  }
}
bool KMP(string ss, int len, string q) {
  int k = -1;
  for (int i = 0; i < lenq; i++) {
    while (k != -1 && ss[k + 1] != q[i]) k = nex[k];
    if (ss[k + 1] == q[i]) k++;
    if (k == len - 1) return true;
  }
  return false;
}
string tmp;
bool pan(string q, int bit) {
  int kk;
  int up = (1 << bit) - 1;
  for (int i = 0; i <= up; i++) {
    tmp.clear();
    for (int j = 0; j < bit; j++) {
      if (i & (1 << j))
        tmp += '1';
      else
        tmp += '0';
    }
    kk = tmp.length();
    GetNext(tmp, kk);
    if (!KMP(tmp, kk, q)) return false;
  }
  return true;
}
int mx;
int solve(string q) {
  lenq = q.length();
  for (int i = mx + 1;; i++) {
    if (!pan(q, i)) {
      printf("%d\n", i - 1);
      return i - 1;
    }
  }
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) cin >> s[i];
  scanf("%d", &m);
  int a, b;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    int lena = s[a].length();
    int lenb = s[b].length();
    p.clear();
    p = s[a] + s[b];
    mx = max(ans[a], ans[b]);
    if (p.length() > 1000)
      p = p.substr(0, 500) + p.substr(lena + lenb - 500, 500);
    s[++n] = p;
    ans[n] = solve(p);
  }
  return 0;
}
