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 N = 110;
const int K = 7;
int n, m;
char s[N][N];
char tmp[N + N];
bool u[N + N][K][1 << K];
char fi[N + N][N], la[N + N][N];
int len[N + N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s[i]);
    for (int j = 0; s[i][j] && j < K; ++j) fi[i][j] = s[i][j];
    int ln = strlen(s[i]);
    int start = ln - K;
    if (start < 0) start = 0;
    while (start < ln) la[i][len[i]++] = s[i][start++];
    for (int ii = 1; ii < K; ++ii)
      for (int j = 0; j < ln - ii + 1; ++j) {
        int t = 0;
        for (int k = j; k < j + ii; ++k) t = t * 2 + s[i][k] - '0';
        u[i][ii][t] = true;
      }
  }
  scanf("%d", &m);
  int num = n - 1;
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x;
    --y;
    ++num;
    for (int i = 1; i < K; ++i)
      for (int j = 0; j < (1 << i); ++j) u[num][i][j] = u[x][i][j] | u[y][i][j];
    int k = 0;
    for (int i = 0; la[x][i]; ++i) tmp[k++] = la[x][i];
    for (int i = 0; fi[y][i]; ++i) tmp[k++] = fi[y][i];
    tmp[k] = 0;
    for (int i = 1; i < K; ++i)
      for (int j = 0; j < k - i + 1; ++j) {
        int t = 0;
        for (int ii = j; ii < j + i; ++ii) t = t * 2 + tmp[ii] - '0';
        u[num][i][t] = true;
      }
    int fk = 0;
    for (fk = 0; fi[x][fk]; ++fk) fi[num][fk] = fi[x][fk];
    for (int i = 0; fi[y][i] && fk < K; ++i) fi[num][fk++] = fi[y][i];
    fi[num][fk] = 0;
    int lk = 0;
    if (len[y] < K) {
      int start = len[x] - K + len[y];
      if (start < 0) start = 0;
      while (start < len[x]) la[num][lk++] = la[x][start++];
    }
    for (int i = 0; la[y][i]; ++i) la[num][lk++] = la[y][i];
    len[num] = lk;
    int r = 0;
    for (int i = K - 1; i >= 0; --i) {
      bool f = true;
      for (int j = 0; j < (1 << i); ++j) f &= u[num][i][j];
      if (f) {
        r = i;
        break;
      }
    }
    printf("%d\n", r);
  }
  return 0;
}