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 M = 13;
struct Str {
  bitset<9010> B[M + 1];
  int pre[20], suf[20];
  int len;
  void init(char* str, int l) {
    len = min(M, l);
    for (int i = 1; i <= len; i++)
      pre[i] = pre[i - 1] << 1 | (str[i - 1] - '0');
    for (int i = 1; i <= len; i++)
      suf[i] = suf[i - 1] | ((str[l - i] - '0') << (i - 1));
    for (int i = 0; i < l; i++) {
      int now = 0;
      for (int j = 1; j <= M && i + j - 1 < l; j++) {
        now = now << 1 | (str[i + j - 1] - '0');
        B[j][now] = 1;
      }
    }
    return;
  }
  int check() {
    for (int i = 1; i <= M; i++)
      for (int j = 0; j < (1 << i); j++)
        if (!B[i][j]) return i - 1;
    return M;
  }
  void output() {
    for (int i = 1; i <= len; i++) printf("%d ", pre[i]);
    printf("\n");
    for (int i = 1; i <= len; i++) printf("%d ", suf[i]);
    printf("\n");
    return;
  }
} S[210];
void merge(Str& a, Str& b, Str& c) {
  for (int i = 1; i <= M; i++) c.B[i] = a.B[i] | b.B[i];
  for (int i = 1; i <= a.len; i++)
    for (int j = 1; j <= b.len; j++)
      if (i + j <= M) c.B[i + j][(a.suf[i] << j) | b.pre[j]] = 1;
  c.len = min(M, a.len + b.len);
  for (int i = 1; i <= c.len; i++)
    if (i <= a.len)
      c.pre[i] = a.pre[i];
    else
      c.pre[i] = (a.pre[a.len] << (i - a.len)) | b.pre[i - a.len];
  for (int i = 1; i <= c.len; i++)
    if (i <= b.len)
      c.suf[i] = b.suf[i];
    else
      c.suf[i] = (a.suf[i - b.len] << b.len) | b.suf[b.len];
  return;
}
char str[110];
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", str);
    int len = strlen(str);
    S[i].init(str, len);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    merge(S[u], S[v], S[n + i]);
    printf("%d\n", S[n + i].check());
  }
  return 0;
}
