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 = 120, mod = int(1e9) + 7;
char last[2 * N][50], first[2 * N][50], second[2 * N][N];
bitset<10000> bit[202][13];
int n, m;
int len[N];
string c;
string cur[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("\n%s", second[i]);
    len[i] = strlen(second[i]);
    for (int j = 0; j < min(len[i], 12); j++) {
      first[i][j] = second[i][j];
    }
    for (int j = max(0, len[i] - 12), first = 0; j < len[i]; first++, j++) {
      last[i][first] = second[i][j];
    }
    for (int j = 0; j < len[i]; j++) {
      int res = 0;
      for (int k = j; k < len[i]; k++) {
        res = res * 2 + (second[i][k] - '0');
        if (k - j + 1 <= 12) {
          int len = k - j + 1;
          bit[i][len][res] = 1;
        } else
          break;
      }
    }
  }
  scanf("%d", &m);
  for (int t = 1, a, b; t <= m; t++) {
    scanf("%d%d", &a, &b);
    int x = strlen(last[a]);
    int y = strlen(first[b]);
    c = "";
    for (int i = 0; i < x; i++) {
      c += last[a][i];
    }
    for (int i = 0; i < y; i++) {
      c += first[b][i];
    }
    for (int i = 1; i <= 12; i++) {
      bit[n + t][i] = (bit[a][i] | bit[b][i]);
    }
    for (int i = 0; i < c.size(); i++) {
      int res = 0;
      for (int j = i; j < c.size(); j++) {
        res = res * 2 + (c[j] - '0');
        int len = j - i + 1;
        if (len <= 12) {
          bit[n + t][len][res] = 1;
        }
      }
    }
    int ans = 0;
    for (int i = 1; i <= 12; i++) {
      if (bit[n + t][i].count() == (1 << i)) {
        ans = i;
      }
    }
    printf("%d\n", ans);
    c = "";
    y = strlen(last[b]);
    x = strlen(last[a]);
    if (y < 12) {
      c = "";
      for (int i = 0; i < x; i++) {
        c += last[a][i];
      }
      for (int i = 0; i < y; i++) {
        c += last[b][i];
      }
      for (int i = max(int(c.size()) - 12, 0), first = 0; i < c.size();
           i++, first++) {
        last[n + t][first] = c[i];
      }
    } else {
      for (int i = 0; i < y; i++) {
        last[n + t][i] = last[b][i];
      }
    }
    x = strlen(first[a]);
    y = strlen(first[b]);
    if (x < 12) {
      for (int i = 0; i < x; i++) {
        first[n + t][i] = first[a][i];
      }
      for (int i = 0; i < y && x + i < 12; i++) {
        first[n + t][x + i] = first[b][i];
      }
    } else {
      for (int i = 0; i < x; i++) {
        first[n + t][i] = first[a][i];
      }
    }
  }
  return 0;
}
