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 MX = 16;
int N, M;
vector<string> S;
struct Info {
  string L, R;
  vector<int> cnt[MX + 1];
};
vector<Info> info;
int main() {
  scanf("%d", &N);
  S.resize(N);
  for (int i = 0; i < N; i++) {
    cin >> S[i];
  }
  scanf("%d", &M);
  info.resize(N + M);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < min((int)S[i].size(), MX); j++)
      info[i].L.push_back(S[i][j]);
    for (int j = S[i].size() - 1; j >= max((int)S[i].size() - MX, 0); j--)
      info[i].R.push_back(S[i][j]);
    reverse(info[i].R.begin(), info[i].R.end());
    for (int j = 1; j <= MX; j++) {
      info[i].cnt[j] = vector<int>((1 << j), 0);
      if (S[i].size() < j) continue;
      int mask = 0;
      for (int k = 0; k < j; k++) {
        if (S[i][k] == '1') mask += (1 << k);
      }
      info[i].cnt[j][mask] = 1;
      for (int k = 1; k <= (int)S[i].size() - j; k++) {
        if (S[i][k - 1] == '1') mask--;
        mask >>= 1;
        if (S[i][k + j - 1] == '1') mask += (1 << (j - 1));
        info[i].cnt[j][mask] = 1;
      }
    }
  }
  for (int i = 0; i < M; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    string L, R;
    vector<int> cnt[MX + 1];
    int pos = 0;
    while (pos < info[a].L.size() && L.size() < MX)
      L.push_back(info[a].L[pos]), pos++;
    pos = 0;
    while (pos < info[b].L.size() && L.size() < MX)
      L.push_back(info[b].L[pos]), pos++;
    pos = info[b].R.size() - 1;
    while (pos >= 0 && R.size() < MX) R.push_back(info[b].R[pos]), pos--;
    pos = info[a].R.size() - 1;
    while (pos >= 0 && R.size() < MX) R.push_back(info[a].R[pos]), pos--;
    reverse(R.begin(), R.end());
    string T = info[a].R + info[b].L;
    for (int j = 1; j <= MX; j++) {
      cnt[j] = vector<int>((1 << j), 0);
      for (int k = 0; k < (1 << j); k++) {
        cnt[j][k] = info[a].cnt[j][k] | info[b].cnt[j][k];
      }
    }
    for (int j = 2; j <= MX; j++) {
      if (T.size() < j) break;
      int mask = 0;
      for (int k = 0; k < j; k++) {
        if (T[k] == '1') mask += (1 << k);
      }
      cnt[j][mask] = 1;
      for (int k = 1; k <= (int)T.size() - j; k++) {
        if (T[k - 1] == '1') mask--;
        mask >>= 1;
        if (T[k + j - 1] == '1') mask += (1 << (j - 1));
        cnt[j][mask] = 1;
      }
    }
    int ans = 0;
    for (int j = MX; j >= 1; j--) {
      bool ok = true;
      for (int k = 0; k < (1 << j); k++) {
        if (!cnt[j][k]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans = j;
        break;
      }
    }
    printf("%d\n", ans);
    info[N + i].L = L;
    info[N + i].R = R;
    for (int j = 0; j <= MX; j++) info[N + i].cnt[j] = cnt[j];
  }
}
