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;
const int MAXN = 100;
const int MAXM = 100;
const int MAXK = 12;
const int INF = 0x3f3f3f3f;
template <class T>
void Read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = 0;
  while (c < '0' || '9' < c) {
    if (c == '-') flag = 1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (flag) x = -x;
}
int n, m;
char origin[MAXM + 10][MAXM + 10];
int a[MAXM + 10], b[MAXM + 10];
char pre[MAXN * 2 + 10][MAXK + 5], suf[MAXN * 2 + 10][MAXK + 5];
bool vis[MAXN * 2 + 10][(1 << MAXK) + 10];
int ans[MAXM + 10];
void getpre(char *pre, char *str) {
  int len = strlen(str + 1);
  for (int i = 1; i <= len && i <= MAXK; ++i) pre[i] = str[i];
  pre[min(len, MAXK) + 1] = '\0';
}
void getsuf(char *suf, char *str) {
  int len = strlen(str + 1);
  int t = min(len, MAXK);
  for (int i = 1; i <= len && i <= MAXK; ++i) suf[t - i + 1] = str[len - i + 1];
  suf[min(len, MAXK) + 1] = '\0';
}
void mark(int id, int k, char *str) {
  int len = strlen(str + 1);
  int t = 0;
  for (int i = 1; i <= len; ++i) {
    if (i > k) t -= (str[i - k] - '0') << (k - 1);
    t = (t << 1) + str[i] - '0';
    if (i >= k) vis[id][t] = true;
  }
}
char str[MAXM + 10];
int merge(int k, int id, int l, int r) {
  int len1 = strlen(suf[l] + 1), len2 = strlen(pre[r] + 1);
  int len = len1 + len2;
  for (int i = 1; i <= len1; ++i) str[i] = suf[l][i];
  for (int i = 1; i <= len2; ++i) str[len1 + i] = pre[r][i];
  str[len + 1] = '\0';
  mark(id, k, str);
  if (len1 >= k)
    memcpy(pre[id], pre[l], sizeof(pre[l]));
  else
    getpre(pre[id], str);
  if (len2 >= k)
    memcpy(suf[id], suf[r], sizeof(suf[r]));
  else
    getsuf(suf[id], str);
  int ed = (1 << k) - 1;
  bool flag = true;
  for (int i = 0; i <= ed; ++i) {
    vis[id][i] |= vis[l][i] | vis[r][i];
    if (!vis[id][i]) flag = false;
  }
  return flag ? k : 0;
}
int main() {
  Read(n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", origin[i] + 1);
    getpre(pre[i], origin[i]), getsuf(suf[i], origin[i]);
  }
  Read(m);
  for (int i = 1; i <= m; ++i) Read(a[i]), Read(b[i]);
  for (int k = 1; k <= MAXK; ++k) {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; ++i) mark(i, k, origin[i]);
    for (int i = 1; i <= m; ++i) {
      ans[i] =
          (merge(k, n + i, a[i], b[i]) == ans[i] + 1 ? ans[i] + 1 : ans[i]);
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}