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;
int n, m, tot, tg[19], s[109];
char ch[109];
struct node {
  int l, r, sl[19], sr[19];
  bool bo[140009];
  void init() {
    scanf("%s", ch + 1);
    int i, j, x, n = strlen(ch + 1);
    for (i = 1; i <= n; i++) s[i] = ch[i] - '0';
    for (i = 1; i <= n; i++) {
      for (j = x = 0; j < 16 && i + j <= n; j++) {
        x = x << 1 | s[i + j];
        bo[tg[j + 1] | x] = 1;
      }
    }
    l = r = min(n, 16);
    for (i = 1; i <= l; i++) sl[i] = s[i];
    for (i = 1; i <= r; i++) sr[i] = s[n - i + 1];
  }
  void mrg(node u, node v) {
    l = u.l;
    memcpy(sl, u.sl, sizeof(u.sl));
    r = v.r;
    memcpy(sr, v.sr, sizeof(v.sr));
    int i, j, x, y;
    for (i = 1; i <= v.l && l < 16; i++) {
      l++;
      sl[l] = v.sl[i];
    }
    for (i = 1; i <= u.r && r < 16; i++) {
      r++;
      sr[r] = u.sr[i];
    }
    for (i = 0; i < tot; i++) bo[i] = u.bo[i] | v.bo[i];
    for (i = 1, x = 0; i <= u.r; i++) {
      x |= u.sr[i] << i - 1;
      for (j = 1, y = x; j <= v.l; j++) {
        y = y << 1 | v.sl[j];
        if (i + j <= 16) bo[tg[i + j] | y] = 1;
      }
    }
  }
  int calc() {
    int i, j, k, x;
    for (i = 1;; i++) {
      x = 1 << i;
      for (j = 0; j < x; j++)
        if (!bo[tg[i] | j]) break;
      if (j < x) break;
    }
    return i - 1;
  }
} a[209];
int main() {
  scanf("%d", &n);
  tot = 1 << 17;
  int i;
  tg[16] = 0;
  for (i = 15; i; i--) tg[i] = tg[i + 1] | 1 << i + 1;
  for (i = 1; i <= n; i++) a[i].init();
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[n + i].mrg(a[x], a[y]);
    printf("%d\n", a[n + i].calc());
  }
  return 0;
}
