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 N = 14;
struct Node {
  Node *L = 0, *R = 0;
};
Node *copy(Node *a) {
  if (!a) return 0;
  Node *r = new Node();
  r->L = copy(a->L);
  r->R = copy(a->R);
  return r;
}
Node *merge(Node *a, Node *b) {
  if (!a) return copy(b);
  if (!b) return copy(a);
  Node *r = new Node();
  r->L = merge(a->L, b->L);
  r->R = merge(a->R, b->R);
  return r;
}
Node *push(Node *a, int pt, int pte, const string &s) {
  if (!a) a = new Node();
  if (pt >= (int)s.size() || pt >= pte) {
    return a;
  }
  if (s[pt] == '0')
    a->L = push(a->L, pt + 1, pte, s);
  else
    a->R = push(a->R, pt + 1, pte, s);
  return a;
}
int cur[N + 1];
void getAns(Node *x, int p) {
  if (!x) return;
  cur[p]++;
  getAns(x->L, p + 1);
  getAns(x->R, p + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> xb, xe;
  vector<Node *> x;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    xb.push_back(s);
    xe.push_back(s);
    x.push_back(new Node());
    for (int i = 0; i < s.size(); ++i) x.back() = push(x.back(), i, i + N, s);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    Node *z = merge(x[a], x[b]);
    xe.push_back(xe[a] + xb[b]);
    for (int j = 0; j < xe[i + n].size(); ++j) {
      z = push(z, j, j + N, xe[i + n]);
    }
    xe[i + n] = xe[a] + xe[b];
    x.push_back(z);
    xb.push_back(xb[a] + xb[b]);
    xb[i + n].resize(min((int)xb[i + n].size(), N));
    if (xe[i + n].size() > N) {
      for (int j = 0; j < N; ++j) {
        xe[i + n][j] = xe[i + n][xe[i + n].size() - N + j];
      }
      xe[i + n].resize(N);
    }
    memset(cur, 0, sizeof cur);
    getAns(z, 0);
    int ans = 0;
    for (int i = 0; i <= N; ++i) {
      if (cur[i] == (1 << i)) ans = i;
    }
    cout << ans << endl;
  }
  return 0;
}