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 M = 210;
const int N = 30;
struct Node {
  int len;
  string l, r;
  set<string> ss;
  int now;
} a[M];
int n;
char s[M];
void update(Node &a) {
  bool flag = true;
  for (int t = a.now + 1;; ++t) {
    for (int j = 0; j < 1 << t && flag; ++j) {
      int v = j;
      string tmp;
      for (int i = 0; i < t; ++i) {
        tmp += '0' + (v >> i & 1);
      }
      flag = a.ss.count(tmp);
    }
    if (!flag) break;
    ++a.now;
  }
  for (auto it = a.ss.begin(); it != a.ss.end();) {
    if ((*it).size() <= a.now)
      it = a.ss.erase(it);
    else
      ++it;
  }
}
void work(Node &a, char *s) {
  int lens = strlen(s);
  a.len = lens;
  for (int i = 0; i < lens && a.l.size() < N; ++i) {
    a.l += s[i];
  }
  for (int i = lens - 1; i >= 0 && a.r.size() < N; --i) {
    a.r += s[i];
  }
  for (int i = 1; i <= lens && i < N; ++i) {
    for (int j = 0; j + i - 1 < lens; ++j) {
      string zz;
      for (int z = 0; z < i; ++z) {
        zz += s[j + z];
      }
      a.ss.insert(zz);
    }
  }
  update(a);
}
void Merge(Node &a, Node &b, Node &c) {
  a.len = b.len + c.len;
  a.l = b.l;
  if (b.len == b.l.size()) {
    for (int j = 0; j < c.l.size() && a.l.size() < N; ++j) {
      a.l += c.l[j];
    }
  }
  a.r = c.r;
  if (c.len == c.r.size()) {
    for (int j = 0; j < b.r.size() && a.r.size() < N; ++j) {
      a.r += b.r[j];
    }
  }
  a.ss = b.ss;
  for (auto &it : c.ss) {
    a.ss.insert(it);
  }
  string d, e;
  a.now = max(b.now, c.now);
  for (int i = 0; i < b.r.size(); ++i) {
    d += b.r[i];
    e = d;
    reverse(e.begin(), e.end());
    for (int j = 0; j < c.l.size() && j < N - (i + 1); ++j) {
      e += c.l[j];
      if (e.size() > a.now) a.ss.insert(e);
    }
  }
  update(a);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s);
    work(a[i], s);
  }
  int Q;
  scanf("%d", &Q);
  for (int j = n; j < n + Q; ++j) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x == y)
      x = y = x - 1;
    else
      --x, --y;
    Merge(a[j], a[x], a[y]);
    printf("%d\n", a[j].now);
  }
  return 0;
}