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 = 100007;
const int inf = (1LL << 29);
int read() {
  int x = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  x *= f;
  return x;
}
int fi[207][10], se[207][10];
int you[207][10][533], l[207];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    char s[107];
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    int x = 1;
    for (int k = 1; k <= min(len, 8); k++) {
      fi[i][k] = fi[i][k - 1] + ((s[k] - '0') << (k - 1));
    }
    int cnt = 0;
    for (int k = len; k >= max(len - 7, 1); k--) {
      se[i][cnt + 1] = (se[i][cnt] << 1) + s[k] - '0';
      ++cnt;
    }
    for (int k = 1; k <= len; k++) {
      int x = 0;
      for (int z = k; z <= min(k + 8, len); z++) {
        x += (s[z] - '0') << (z - k);
        you[i][z - k + 1][x] = 1;
      }
    }
    l[i] = len;
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    l[n + i] = min(100, l[a] + l[b]);
    for (int k = 1; k <= 9; k++)
      for (int z = 0; z < (1 << k); z++) {
        if (you[a][k][z] || you[b][k][z]) you[n + i][k][z] = 1;
      }
    for (int k = 1; k <= min(8, l[a] + l[b]); k++) {
      if (k <= l[a])
        fi[n + i][k] = fi[a][k];
      else
        fi[n + i][k] = fi[a][l[a]] + (fi[b][k - l[a]] << l[a]);
    }
    for (int k = 1; k <= min(8, l[a] + l[b]); k++) {
      if (k <= l[b])
        se[n + i][k] = se[b][k];
      else
        se[n + i][k] = se[a][k - l[b]] + (se[b][l[b]] << (k - l[b]));
    }
    for (int k = 1; k <= min(l[a], 8); k++)
      for (int z = 1; z <= min(l[b], 9 - k); z++) {
        you[n + i][k + z][se[a][k] + (fi[b][z] << k)] = 1;
      }
    int ans = 0;
    for (int k = 1; k <= 9; k++) {
      int ok = 1;
      for (int z = 0; z < (1 << k); z++) {
        if (!you[n + i][k][z]) ok = 0;
      }
      if (ok) ans = k;
    }
    printf("%d\n", ans);
  }
  return 0;
}