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 double eps = 1e-6;
const int inf = 1e9;
const int mx = 13;
unsigned long long read() {
  unsigned long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
char s[1005];
struct str {
  char L[55], R[55];
  int len;
  bitset<8211> b[13];
  void read() {
    for (int i = 0; i <= mx - 1; i++) b[i].reset();
    scanf("%s", s);
    len = strlen(s);
    for (int i = 0; i <= len - 1; i++) {
      int v = 0;
      for (int j = 0; j < mx && i + j < len; j++)
        v = v << 1 | (s[i + j] - '0'), b[j].set(v);
    }
    for (int i = 0; i < len && i < mx; i++) L[i] = s[i], R[i] = s[len - 1 - i];
    if (len > mx) len = mx;
  }
  str operator+(const str& w) {
    str r;
    r.len = min(len + w.len, mx);
    for (int i = 0; i <= mx - 1; i++) r.b[i] = b[i] | w.b[i];
    for (int i = 0; i <= len - 1; i++) r.L[i] = L[i];
    for (int i = 0; i <= w.len - 1; i++) r.L[i + len] = w.L[i];
    for (int i = 0; i <= w.len - 1; i++) r.R[i] = w.R[i];
    for (int i = 0; i <= len - 1; i++) r.R[i + w.len] = R[i];
    for (int i = 0; i <= len - 1; i++) {
      int v = 0;
      for (int j = i; j >= 0; j--) v = v << 1 | (R[j] - '0');
      for (int j = 0; j < w.len && i + j + 1 < mx; j++) {
        v = v << 1 | (w.L[j] - '0');
        r.b[i + j + 1].set(v);
      }
    }
    return r;
  }
  void pr() {
    for (int i = 0; i <= mx - 1; i++) {
      if (b[i].count() < (2u << i)) {
        printf("%d\n", i);
        return;
      }
    }
  }
} ss[205];
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) ss[i].read();
  int m = read();
  for (int i = 0, a, b; i < m; i++) {
    a = read();
    b = read();
    ss[++n] = ss[a] + ss[b];
    ss[n].pr();
  }
  return 0;
}