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;
set<string> st[400][25];
string a[400][2];
int n, x, y, Q;
char s[401];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    int m = strlen(s);
    a[i][0] = "";
    for (int j = 0; j < m; j++) {
      string now = "";
      int len = 0;
      if (a[i][0].length() < 21) a[i][0] = a[i][0] + s[j];
      for (int k = j; k < m; k++) {
        len++;
        now = now + s[k];
        st[i][len].insert(now);
        if (len > 20) break;
      }
    }
    for (int j = max(0, m - 21); j < m; j++) a[i][1] = a[i][1] + s[j];
  }
  scanf("%d", &Q);
  for (int i = n + 1; i <= n + Q; i++) {
    scanf("%d%d", &x, &y);
    for (int j = 1; j <= 21; j++) st[i][j] = st[x][j];
    for (int j = 0; j < a[x][1].length(); j++)
      for (int k = 0; k < a[y][0].length(); k++) {
        string now = "";
        now = now + a[x][1].substr(j, a[x][1].length() - j);
        now = now + a[y][0].substr(0, k + 1);
        st[i][now.length()].insert(now);
        if (now.length() > 20) break;
      }
    for (int len = 1; len <= 21; len++)
      for (set<string>::iterator it = st[y][len].begin();
           it != st[y][len].end(); it++)
        st[i][len].insert(*it);
    int ans = 0;
    for (int len = 1; len <= 21; len++) {
      if (st[i][len].size() == (1 << len)) ans = len;
    }
    printf("%d\n", ans);
    a[i][0] = a[x][0];
    a[i][1] = a[y][1];
    if (a[i][0].length() < 21) {
      for (int j = 0; j < a[y][0].length(); j++) {
        a[i][0] += a[y][0][j];
        if (a[i][0].length() > 20) break;
      }
    }
    string now = "";
    if (a[i][1].length() < 21) {
      for (int j = a[x][1].length() - 1; j >= 0; j--) {
        now += a[x][1][j];
        if (now.length() + a[i][1].length() > 20) break;
      }
    }
    reverse(now.begin(), now.end());
    a[i][1] = now + a[i][1];
  }
  return 0;
}