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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int getid(string s) {
  int len = ((int)(s).size()), id = 0;
  if (len > 19) return -1;
  for (int j = 0; j < len; j++) id = id * 2 + s[j] - '0';
  return id + (1 << len) - 2;
}
int n, m, u, v;
char s[110];
string t;
struct blk {
  string bg, ed;
  bitset<1048576> S;
  bool full;
  void add(string s) {
    int m = ((int)(s).size());
    for (int j = 0; j < m; j++)
      for (int k = j; k < m; k++) {
        string f = s.substr(j, k - j + 1);
        int id = getid(f);
        if (id == -1) continue;
        S[id] = 1;
      }
  }
  void simply() {
    if (((int)(bg).size()) > 20) {
      bg = bg.substr(0, 20);
      full = 0;
    }
    if (((int)(ed).size()) > 20) {
      ed = ed.substr(((int)(ed).size()) - 20, 20);
      full = 0;
    }
  }
  int query() {
    for (int i = 20 - 1; i >= 1; i--) {
      int d = (1 << i) - 2, val = 1;
      for (int j = 0; j < (1 << i); j++) {
        if (!S[d + j]) {
          val = 0;
          break;
        }
      }
      if (val) return i;
    }
    return 0;
  }
} b[210];
void gao(int p, int u, int v) {
  b[p].S = b[u].S | b[v].S;
  if (b[u].full)
    b[p].bg = b[u].bg + b[v].bg, b[p].full = 1;
  else
    b[p].bg = b[u].bg;
  if (b[v].full)
    b[p].ed = b[u].ed + b[v].ed, b[p].full = 1;
  else
    b[p].ed = b[v].ed;
  b[p].add(b[u].ed + b[v].bg);
  b[p].simply();
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    t = s;
    b[i].add(t);
    b[i].bg = t;
    b[i].ed = t;
    b[i].full = 1;
    b[i].simply();
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    gao(n + i, u - 1, v - 1);
    printf("%d\n", b[n + i].query());
  }
}