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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int maxm = 210;
long long ans, K;
struct SAM {
  int ch[maxn][2];
  int pre[maxn], step[maxn];
  int last, id;
  int nu[maxn];
  void init() {
    last = id = 0;
    memset(ch[0], -1, sizeof(ch[0]));
    pre[0] = -1;
    step[0] = 0;
  }
  void insert(int c) {
    int p = last, np = ++id;
    step[np] = step[p] + 1;
    memset(ch[np], -1, sizeof(ch[np]));
    while (p != -1 && ch[p][c] == -1) ch[p][c] = np, p = pre[p];
    if (p == -1)
      pre[np] = 0;
    else {
      int q = ch[p][c];
      if (step[q] != step[p] + 1) {
        int nq = ++id;
        memcpy(ch[nq], ch[q], sizeof(ch[q]));
        step[nq] = step[p] + 1;
        pre[nq] = pre[q];
        pre[np] = pre[q] = nq;
        while (p != -1 && ch[p][c] == q) ch[p][c] = nq, p = pre[p];
      } else
        pre[np] = q;
    }
    last = np;
  }
} sam;
unordered_map<string, bool> sk;
bitset<64> ss;
bool judge(const int &mid) {
  sk.clear();
  long long maxx = min(1LL << (mid + 5), 10000LL);
  for (int i = 0; i < maxx; i++) {
    ss = rand();
    string now = ss.to_string();
    if (sk[now]) continue;
    sk[now] = true;
    int id = 0;
    for (int j = 0; j < mid; j++) {
      if (sam.ch[id][ss[j]] == -1) return false;
      id = sam.ch[id][ss[j]];
    }
  }
  return true;
}
string s[maxm];
int n, m;
int dp[maxm];
int solve(int id) {
  sam.init();
  int len = s[id].length();
  for (int i = 0; i < len; i++) sam.insert(s[id][i] - '0');
  int low = 0, high = 20;
  while (true) {
    int mid = (low + high) >> 1;
    if (mid == low) {
      if (judge(high)) low = high;
      break;
    }
    if (judge(mid))
      low = mid;
    else
      high = mid - 1;
  }
  return low;
}
int main() {
  srand((unsigned)time(NULL));
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> s[i];
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int a, b;
    cin >> a >> b;
    s[i] = s[a] + s[b];
    if (s[i].length() > 1000)
      s[i] = s[i].substr(0, 500) + s[i].substr(s[i].length() - 500, 500);
    dp[i] = max({dp[a], dp[b], solve(i)});
    cout << dp[i] << endl;
  }
  return 0;
}
