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;
string a[210][2];
int lc[210], rc[210];
int vis[210];
int len[210];
map<string, int> mp;
long long cnt;
void dfs(int now, int k) {
  if (vis[now]) return;
  vis[now] = 1;
  if (lc[now] == 0) {
    for (int i = 0; i + k <= a[now][0].length(); i++) {
      string t = a[now][0].substr(i, k);
      if (mp[t] == 0) cnt++;
      mp[t]++;
    }
  } else {
    string s = a[lc[now]][1] + a[rc[now]][0];
    for (int i = 0; i + k <= s.length(); i++) {
      string t = s.substr(i, k);
      if (mp[t] == 0) cnt++;
      mp[t]++;
    }
  }
  dfs(lc[now], k);
  dfs(rc[now], k);
}
int check(int now, int k) {
  mp.clear();
  memset(vis, 0, sizeof(vis));
  cnt = 0;
  dfs(now, k);
  if (cnt == (1 << k))
    return 1;
  else
    return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i][0];
    a[i][1] = a[i][0];
    lc[i] = rc[i] = 0;
    len[i] = a[i][0].length();
  }
  int q;
  scanf("%d", &q);
  for (int i = n + 1; i <= n + q; i++) {
    int zuo, you;
    scanf("%d%d", &zuo, &you);
    lc[i] = zuo;
    rc[i] = you;
    a[i][0] = a[lc[i]][0];
    if (len[lc[i]] <= 100) a[i][0] = a[i][0] + a[rc[i]][0];
    if (a[i][0].length() >= 100) a[i][0] = a[i][0].substr(0, 100);
    a[i][1] = a[rc[i]][1];
    if (len[rc[i]] <= 100) a[i][1] = a[lc[i]][1] + a[i][1];
    if (a[i][1].length() >= 100)
      a[i][1] = a[i][1].substr(a[i][1].length() - 100, 100);
    int l = 0, r = 100;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (check(i, mid))
        l = mid + 1;
      else
        r = mid - 1;
    }
    printf("%d\n", r);
  }
}