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 L = 1 << 17;
set<int> X[205];
string A[205], B[205];
int main() {
  int N, M;
  scanf("%d", &N);
  for (int t = 1; t <= N; t++) {
    string P;
    cin >> P;
    for (int i = 0; i < P.size(); i++) {
      int k = 1;
      for (int j = i; j < P.size(); j++) {
        k = k * 2 + P[j] - '0';
        if (k >= L) break;
        X[t].insert(k);
      }
    }
    A[t] = P.substr(0, min(16, (int)P.size()));
    B[t] = P.substr(max(0, (int)P.size() - 16), -1);
  }
  scanf("%d", &M);
  for (int t = N + 1; t <= M + N; t++) {
    int a, b;
    scanf("%d%d", &a, &b);
    string P = B[a], Q = A[b], R = B[a] + A[b];
    for (int i = 0; i < R.size(); i++) {
      int k = 1;
      for (int j = i; j < R.size(); j++) {
        k = k * 2 + R[j] - '0';
        if (k >= L) break;
        X[t].insert(k);
      }
    }
    for (int c : X[a]) X[t].insert(c);
    for (int c : X[b]) X[t].insert(c);
    if (P.size() <= 16) {
      string tt = A[a] + A[b];
      A[t] = tt.substr(0, min(16, (int)tt.size()));
    }
    if (Q.size() <= 16) {
      string tt = B[a] + B[b];
      B[t] = tt.substr(max(0, (int)tt.size() - 16), -1);
    }
  }
  for (int i = N + 1; i <= N + M; i++) {
    auto it = X[i].begin();
    int t = 2;
    for (t = 2; it != X[i].end(); t++, it++) {
      if (*it != t) break;
    }
    int ans = 0;
    while (t >= 1 << ans + 2) ans++;
    printf("%d\n", ans);
  }
}