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 N = 505;
int n, m, cnt;
bool vis[N];
int lch[N], rch[N];
string a[N][2];
set<string> dic;
void dfs(int i, int k) {
  if (vis[i]) return;
  vis[i] = true;
  string s, t;
  if (!lch[i])
    s = a[i][0];
  else
    s = a[lch[i]][1] + a[rch[i]][0];
  for (int i = 0; i + k <= s.length(); i++) {
    t = s.substr(i, k);
    if (!dic.count(t)) {
      dic.insert(t);
      cnt++;
    }
  }
  if (lch[i]) dfs(lch[i], k);
  if (rch[i]) dfs(rch[i], k);
}
bool check(int k, int i) {
  memset(vis, false, sizeof(vis));
  dic.clear();
  cnt = 0;
  dfs(i, k);
  return cnt == (1 << k);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i][0];
    a[i][1] = a[i][0];
    lch[i] = rch[i] = 0;
  }
  scanf("%d", &m);
  for (int i = n + 1; i <= n + m; i++) {
    scanf("%d%d", lch + i, rch + i);
    a[i][0] = a[lch[i]][0];
    if (a[i][0].length() < 100) a[i][0] = a[i][0] + a[rch[i]][0];
    if (a[i][0].length() > 100) a[i][0] = a[i][0].substr(0, 100);
    a[i][1] = a[rch[i]][1];
    if (a[i][1].length() < 100) a[i][1] = a[lch[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, mid;
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (check(mid, i))
        l = mid;
      else
        r = mid - 1;
    }
    printf("%d\n", l);
  }
}
