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 N = 205;
vector<bool> g[N][11];
string s[N], L[N], R[N];
char arr[N];
void pre(int in) {
  for (int i = 1; i < 10; i++) {
    vector<bool> temp(1 << i);
    for (int j = 0, x = 0; j + i <= s[in].size(); j++) {
      string d = s[in].substr(j, i);
      x = 0;
      for (int k = 0; k < i; k++) {
        x *= 2;
        x += d[k] - '0';
      }
      temp[x] = 1;
    }
    g[in][i] = temp;
  }
  if (s[in].size() > 20) {
    L[in] = s[in].substr(0, 10);
    R[in] = s[in].substr((int)s[in].size() - 10, 10);
  } else
    L[in] = R[in] = s[in];
}
string emp = "                         ";
int add(int in, int a, int b) {
  int ret = 0;
  for (int i = 1; i < 10; i++) {
    int am = (1 << i);
    vector<bool> temp(1 << i);
    for (int j = 0; j < (1 << i); j++) {
      temp[j] = g[a][i][j] | g[b][i][j];
      if (temp[j]) am--;
    }
    string S = R[a] + L[b];
    for (int j = 0, x = 0; j + i <= S.size(); j++) {
      string d = S.substr(j, i);
      x = 0;
      for (int k = 0; k < i; k++) {
        x *= 2;
        x += d[k] - '0';
      }
      if (!temp[x]) am--;
      temp[x] = 1;
    }
    if (s[a].size() + s[b].size() <= 20) {
      s[in] = s[a] + s[b];
      L[in] = R[in] = s[in];
    } else {
      s[in] = emp;
      if (L[a].size() >= 10)
        L[in] = L[a];
      else {
        if (L[a].size() + L[b].size() < 10)
          L[in] = L[a] + L[b];
        else
          L[in] = L[a] + L[b].substr(0, 10 - L[a].size());
      }
      if (R[b].size() >= 10)
        R[in] = R[b];
      else {
        if (R[b].size() + R[a].size() < 10)
          R[in] = R[a] + R[b];
        else {
          int len = 10 - R[b].size();
          R[in] = R[a].substr(R[a].size() - len, len) + R[b];
        }
      }
    }
    if (!am) ret = i;
    g[in][i] = temp;
  }
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", arr);
    s[i] = arr;
    pre(i);
  }
  int q;
  scanf("%d", &q);
  for (int in = n, a, b; in < n + q; in++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    printf("%d\n", add(in, a, b));
  }
  return 0;
}