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;
struct str {
  string pre, suf;
  int r;
  char a[65536];
} d[202];
str operator+(const str &u, const str &v) {
  str res;
  int i, j, t, tt;
  for (i = 0; i < 65536; i++) res.a[i] = u.a[i] || v.a[i];
  if (u.pre == u.suf) {
    res.pre = u.pre + v.pre;
    if (res.pre.size() > 20) res.pre.resize(20);
  } else
    res.pre = u.pre;
  if (v.pre == v.suf) {
    res.suf = u.suf + v.suf;
    if (res.suf.size() > 20) res.suf = res.suf.substr(res.suf.size() - 20, 20);
  } else
    res.suf = v.suf;
  t = 0;
  for (i = 0; i < 16 && i < u.suf.size(); i++) {
    t |= (u.suf[u.suf.size() - i - 1] & 1) << i;
    tt = t;
    for (j = 0; j < 16 && j < v.pre.size(); j++) {
      tt = tt << 1 | v.pre[j] & 1;
      if (i + j + 2 < 16) res.a[1 << i + j + 2 | tt] = 1;
    }
  }
  for (res.r = 0;; res.r++) {
    for (i = 0; i < 1 << res.r && res.a[1 << res.r | i]; i++)
      ;
    if (i < 1 << res.r) break;
  }
  return res;
}
char tt[256];
int main() {
  int i, j, k, n, m, t;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%s", tt);
    d[i].pre = d[i].suf = tt;
    d[i].a[1] = 1;
    for (j = 0; j < d[i].pre.size(); j++)
      for (k = t = 0; k < 15 && j + k < d[i].pre.size(); k++) {
        t = t << 1 | d[i].pre[j + k] & 1;
        d[i].a[1 << k + 1 | t] = 1;
      }
  }
  scanf("%d", &m);
  for (k = 1; k <= m; k++) {
    scanf("%d%d", &i, &j);
    d[n + k] = d[i] + d[j];
    printf("%d\n", d[n + k].r - 1);
  }
  return 0;
}