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;
int n, m;
int lim = 16;
char s[210];
char sl[210][30], sr[210][30];
int lenl[210], lenr[210];
bool has[210][17][66536];
int K[210];
int rem[210][17];
long long T;
int main() {
  int i, j, k, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    for (j = 1; j <= lim && j <= l; j++) {
      sl[i][++lenl[i]] = s[j];
    }
    for (j = max(1, l - lim + 1); j <= l; j++) {
      sr[i][++lenr[i]] = s[j];
    }
    for (k = 1; k <= lim; k++) {
      rem[i][k] = 1 << k;
    }
    for (k = 1; k <= lim && k <= l; k++) {
      int now = 0;
      for (j = 1; j < k; j++) {
        now <<= 1;
        now |= s[j] - '0';
      }
      for (; j <= l; j++) {
        now <<= 1;
        now |= s[j] - '0';
        now &= (1 << k) - 1;
        if (!has[i][k][now]) {
          rem[i][k]--;
        }
        has[i][k][now] = 1;
      }
    }
  }
  scanf("%d", &m);
  while (m--) {
    n++;
    scanf("%d%d", &x, &y);
    for (i = 1; i <= lenl[x]; i++) {
      s[i] = sl[x][i];
    }
    for (i = 1; i <= lenl[y]; i++) {
      s[lenl[x] + i] = sl[y][i];
    }
    for (i = 1; i <= lenl[x] + lenl[y] && i <= lim; i++) {
      sl[n][++lenl[n]] = s[i];
    }
    int t = 0;
    for (i = lenr[y]; i; i--) {
      s[++t] = sr[y][i];
    }
    for (i = lenr[x]; i; i--) {
      s[++t] = sr[x][i];
    }
    for (i = 1; i <= lenr[x] + lenr[y] && i <= lim; i++) {
      sr[n][++lenr[n]] = s[i];
    }
    for (i = 1; i <= lenr[n] / 2; i++) {
      swap(sr[n][i], sr[n][lenr[n] - i + 1]);
    }
    for (i = 1; i <= lenr[x]; i++) {
      s[i] = sr[x][i];
    }
    for (i = 1; i <= lenl[y]; i++) {
      s[lenr[x] + i] = sl[y][i];
    }
    for (k = 1; k <= lim; k++) {
      rem[n][k] = 1 << k;
      for (j = 0; j < (1 << k); j++) {
        T++;
        has[n][k][j] = has[x][k][j] | has[y][k][j];
        if (has[n][k][j]) {
          rem[n][k]--;
        }
      }
      if (k <= lenr[x] + lenl[y]) {
        int now = 0;
        for (j = 1; j < k; j++) {
          now <<= 1;
          now |= s[j] - '0';
        }
        for (; j <= lenr[x] + lenl[y]; j++) {
          T++;
          now <<= 1;
          now |= s[j] - '0';
          now &= (1 << k) - 1;
          if (!has[n][k][now]) {
            rem[n][k]--;
          }
          has[n][k][now] = 1;
        }
      }
    }
    int ans = 0;
    for (i = 1; i <= lim; i++) {
      if (!rem[n][i]) {
        ans = i;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}