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 MAXN = 202;
const int INF = 0x3f3f3f3f;
int n, m, ans;
char s[MAXN][MAXN];
int len[MAXN];
set<int> st[MAXN][11];
set<int>::iterator it;
int pre[MAXN][11], suf[MAXN][11];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i]);
    len[i] = strlen(s[i]);
    for (int j = 0; j < min(10, len[i]); j++) {
      pre[i][j] = s[i][j] - 48;
    }
    for (int j = 0; j < min(10, len[i]); j++) {
      suf[i][j] = s[i][len[i] - j - 1] - 48;
    }
    for (int j = 1; j <= 10 && j <= len[i]; j++) {
      int tmp = 0;
      for (int k = 0; k < j - 1; k++) tmp = tmp * 2 + s[i][k] - 48;
      for (int k = j - 1; k < len[i]; k++) {
        tmp = (tmp * 2 + s[i][k] - 48) & ((1 << j) - 1);
        st[i][j].insert(tmp);
      }
    }
  }
  scanf("%d", &m);
  while (m--) {
    n++;
    int x, y;
    scanf("%d%d", &x, &y);
    len[n] = min(len[x] + len[y], INF);
    for (int i = 0; i < min(10, len[x]); i++) pre[n][i] = pre[x][i];
    for (int i = min(10, len[x]); i < 10 && i < len[n]; i++)
      pre[n][i] = pre[y][i - len[x]];
    for (int i = 0; i < min(10, len[y]); i++) suf[n][i] = suf[y][i];
    for (int i = min(10, len[y]); i < 10 && i < len[n]; i++)
      suf[n][i] = suf[x][i - len[y]];
    for (int i = 1; i <= 10; i++) {
      for (it = st[x][i].begin(); it != st[x][i].end(); ++it) {
        st[n][i].insert(*it);
      }
      for (it = st[y][i].begin(); it != st[y][i].end(); ++it) {
        st[n][i].insert(*it);
      }
    }
    for (int i = 1; i <= min(10, len[x]); i++) {
      for (int j = 1; j <= min(10, len[y]); j++)
        if (i + j <= 10) {
          int tmp = 0;
          for (int k = i - 1; k >= 0; k--) {
            tmp = tmp * 2 + suf[x][k];
          }
          for (int k = 0; k < j; k++) {
            tmp = tmp * 2 + pre[y][k];
          }
          st[n][i + j].insert(tmp);
        }
    }
    ans = 0;
    for (int i = 10; i; i--)
      if (st[n][i].size() == (unsigned)(1 << i)) {
        ans = i;
        break;
      }
    printf("%d\n", ans);
  }
  return 0;
}