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 MAXK = 12;
const int MAXN = 100 + 3;
const int MAXS = 1 << (MAXK + 2);
int N, M;
string s[MAXN * 2][2];
bool ok[MAXN * 2][MAXK + 3][MAXS];
void work(int id, const string &s, int length) {
  for (int len = 1; len <= min(MAXK, length); ++len)
    for (int l = 0; l + len <= length; ++l) {
      int x = 0;
      for (int r = l; r - l < len; ++r) {
        x <<= 1;
        x |= s[r] == '1';
      }
      ok[id][len][x] = true;
    }
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    char tmp[MAXN];
    scanf("%s", tmp);
    int len0 = strlen(tmp);
    int len = min(len0, MAXK);
    for (int j = 0; j < len; ++j) s[i][0] += tmp[j];
    for (int j = len; j > 0; --j) s[i][1] += tmp[len0 - j];
    work(i, tmp, strlen(tmp));
  }
  scanf("%d", &M);
  for (int i = 0; i < M; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    string comb = s[a][1] + s[b][0];
    for (int k = 1; k <= MAXK; ++k)
      for (int s = 0; s < MAXS; ++s)
        ok[N + i][k][s] = ok[a][k][s] | ok[b][k][s];
    work(N + i, comb, comb.size());
    s[N + i][0] = s[a][0];
    if (s[N + i][0].size() < MAXK) {
      int len = min((int)comb.size(), MAXK);
      s[N + i][0] = "";
      for (int j = 0; j < len; ++j) s[N + i][0] += comb[j];
    }
    s[N + i][1] = s[b][1];
    if (s[N + i][1].size() < MAXK) {
      int len = min((int)comb.size(), MAXK);
      s[N + i][1] = "";
      for (int j = len; j > 0; --j) s[N + i][1] += comb[comb.size() - j];
    }
    bool find = false;
    for (int len = MAXK; len > 0; --len) {
      bool flag = true;
      int S = (1 << len);
      for (int s = 0; s < S; ++s)
        if (!ok[N + i][len][s]) {
          flag = false;
          break;
        }
      if (flag) {
        printf("%d\n", len);
        find = true;
        break;
      }
    }
    if (!find) puts("0");
  }
  return 0;
}