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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
int IT_MAX = 1 << 17;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x1f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
char u[105];
string in[205][2];
set<string> Sx[205][14];
int main() {
  int N, M, i, j, k;
  int mx = 13;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%s", u);
    int L = strlen(u);
    for (j = 0; j < L; j++) {
      for (k = j; k < L; k++) {
        if (k - j + 1 > mx) break;
        Sx[i][k - j + 1].emplace(u + j, u + k + 1);
      }
    }
    if (L <= mx) {
      in[i][0] = string(u);
      in[i][1] = string(u);
    } else {
      in[i][0] = string(u, u + mx);
      in[i][1] = string(u + L - mx, u + L);
    }
  }
  scanf("%d", &M);
  for (i = N + 1; i <= M + N; i++) {
    int t1, t2;
    scanf("%d %d", &t1, &t2);
    for (j = 1; j <= mx; j++) {
      for (auto it : Sx[t1][j]) Sx[i][j].insert(it);
      for (auto it : Sx[t2][j]) Sx[i][j].insert(it);
    }
    string str = in[t1][1] + in[t2][0];
    for (j = 0; j < str.size(); j++) {
      for (k = j; k < str.size(); k++) {
        if (k - j + 1 > mx) break;
        Sx[i][k - j + 1].emplace(str.begin() + j, str.begin() + k + 1);
      }
    }
    for (j = 1; j <= mx; j++)
      if (Sx[i][j].size() != (1 << j)) break;
    printf("%d\n", j - 1);
    if (in[t1][0].size() == mx)
      in[i][0] = in[t1][0];
    else {
      string str = in[t1][0] + in[t2][0];
      if (str.size() <= mx)
        in[i][0] = str;
      else
        in[i][0] = string(str.begin(), str.begin() + mx);
    }
    if (in[t2][1].size() == mx)
      in[i][1] = in[t2][1];
    else {
      string str = in[t1][1] + in[t2][1];
      if (str.size() <= mx)
        in[i][1] = str;
      else
        in[i][1] =
            string(str.begin() + str.size() - mx, str.begin() + str.size());
    }
  }
  return 0;
}
