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 MXN = 2e2 + 10, N = 10, MK = (1 << N);
int n, m, r, t, a, b, Mn, P[MXN][MK], S[MXN][MK], A[MXN][MK], Vl[MXN], sz[MXN];
char T[MXN];
int F(int x) {
  if (x == 0) return (0);
  return (floor(log2(x)));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 205; i++)
    for (int j = 0; j < MK; j++) P[i][j] = S[i][j] = A[i][j] = F(j);
  for (int i = 1; i <= n; i++) {
    scanf("%s", &T);
    r = sz[i] = strlen(T);
    for (int j = 0; j < r; j++) {
      t = 0;
      for (int h = j; h < r; h++) {
        t = t * 2 + (T[h] == '1');
        if (t >= MK) break;
        A[i][t] = max(A[i][t], h - j + 1);
        if (j == 0) P[i][t] = max(P[i][t], h - j + 1);
        if (h == r - 1) S[i][t] = max(S[i][t], h - j + 1);
      }
    }
    if (r > N) {
      Vl[i] = -1;
      continue;
    }
    t = 0;
    for (int j = 0; j < r; j++) t = t * 2 + (T[j] == '1');
    Vl[i] = t;
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    sz[n + i] = min(sz[a] + sz[b], MK * MK);
    if (Vl[a] == -1 || Vl[b] == -1 || Vl[a] * (1 << sz[b]) + Vl[b] >= MK)
      Vl[n + i] = -1;
    else
      Vl[n + i] = Vl[a] * (1 << sz[b]) + Vl[b];
    for (int j = 0; j < MK; j++) A[n + i][j] = max(A[a][j], A[b][j]);
    for (int j = 0; j < MK; j++) P[n + i][j] = P[a][j], S[n + i][j] = S[b][j];
    for (int j = 0; j < MK; j++) {
      if (S[a][j] == F(j)) continue;
      for (int h = 0; h < MK; h++) {
        if (P[b][h] == F(h)) continue;
        if (P[b][h] > N) continue;
        r = j * (1 << P[b][h]) + h;
        if (r < MK) A[n + i][r] = max(A[n + i][r], S[a][j] + P[b][h]);
      }
      for (int h = 1; h <= P[b][0]; h++) {
        if (h > N) continue;
        r = j * (1 << h);
        if (r < MK) A[n + i][r] = max(A[n + i][r], S[a][j] + h);
      }
    }
    if (Vl[a] != -1) {
      for (int j = 0; j < MK; j++) {
        if (P[b][j] == F(j)) continue;
        if (P[b][j] > N) continue;
        r = Vl[a] * (1 << P[b][j]) + j;
        if (r < MK) P[n + i][r] = max(P[n + i][r], sz[a] + P[b][j]);
      }
      for (int h = 1; h <= P[b][0]; h++) {
        if (h > N) continue;
        r = Vl[a] * (1 << h);
        if (r < MK) P[n + i][r] = max(P[n + i][r], sz[a] + h);
      }
    }
    if (Vl[b] != -1) {
      for (int j = 0; j < MK; j++) {
        if (S[a][j] == F(j)) continue;
        if (sz[b] > N) continue;
        r = j * (1 << sz[b]) + Vl[b];
        if (r < MK) S[n + i][r] = max(S[n + i][r], sz[b] + S[a][j]);
      }
    }
    if (Vl[n + i] != -1) {
      A[n + i][Vl[n + i]] = max(A[n + i][Vl[n + i]], sz[n + i]);
      P[n + i][Vl[n + i]] = max(P[n + i][Vl[n + i]], sz[n + i]);
      S[n + i][Vl[n + i]] = max(S[n + i][Vl[n + i]], sz[n + i]);
    }
    for (int j = 0; j < MK; j++)
      A[n + i][j] = max(A[n + i][j], max(P[n + i][j], S[n + i][j]));
    Mn = 1e9;
    for (int j = 0; j < MK; j++) Mn = min(Mn, A[n + i][j]);
    printf("%d\n", Mn);
  }
  return (0);
}