You are given n strings s_1, s_2, ..., s_{n} 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 s_{a}_{i}s_{b}_{i} is saved into a new string s_{n} + 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 2^{k} 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 s_1, s_2, ..., s_{n} (1 ≤ |s_{i}| ≤ 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 a_{i} abd b_{i} (1 ≤ a_{i}, b_{i} ≤ n + i - 1) — the number of strings that are concatenated to form s_{n} + 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.
# -*- coding: utf-8 -*-

import math
import collections
import bisect
import heapq
import time
import random

"""
created by shhuan at 2017/10/5 15:00

"""

N = int(input())
S = ['']
for i in range(N):
    S.append(input())
M = int(input())

# t0 = time.time()
# N = 3
# S = ["", "00010110000", "110101110101101010101101101110100010000001101101011000010001011000010101", "11100101100111010"]
# M = 1000

A = [[set() for _ in range(10)] for _ in range(M+N+1)]
D = collections.defaultdict(int)

for i in range(1, N+1):
    for j in range(1, 10):
        s = S[i]
        if j > len(s):
            break
        for k in range(len(s)-j+1):
            A[i][j].add(int(s[k:k+j], 2))

        if all(v in A[i][j] for v in range(2**j)):
            D[i] = j

for i in range(M):
    # a, b = random.randint(1, i+N), random.randint(1, i+N)
    a, b = list(map(int, input().split()))
    s, sa, sb = S[a] + S[b], S[a], S[b]
    if len(s) > 30:
        S.append(s[:10] + s[-10:])
    else:
        S.append(s)

    ai = i+N+1

    d = max(D[a], D[b]) + 1
    for dv in range(d, 10):
        if len(sa) + len(sb) < dv:
            break
        A[ai][dv] = A[a][dv] | A[b][dv] | {int(v, 2) for v in {sa[-i:] + sb[:dv-i] for i in range(1, dv+1)} if len(v) == dv}

    ans = d-1
    for dv in range(d, 10):
        if any(v not in A[ai][dv] for v in range(2**dv)):
            break
        ans = dv
    print(ans)
    D[ai] = ans





# print(time.time() - t0)





# -*- coding: utf-8 -*-

import math
import collections
import bisect
import heapq
import time
import random

"""
created by shhuan at 2017/10/5 15:00

"""

N = int(input())
S = ['']
for i in range(N):
    S.append(input())
M = int(input())

# t0 = time.time()
# N = 3
# S = ["", "00010110000", "110101110101101010101101101110100010000001101101011000010001011000010101", "11100101100111010"]
# M = 1000

A = [[set() for _ in range(10)] for _ in range(M+N+1)]
D = collections.defaultdict(int)

for i in range(1, N+1):
    for j in range(1, 10):
        s = S[i]
        if j > len(s):
            break
        for k in range(len(s)-j+1):
            A[i][j].add(int(s[k:k+j], 2))

        if all(v in A[i][j] for v in range(2**j)):
            D[i] = j

for i in range(M):
    # a, b = random.randint(1, i+N), random.randint(1, i+N)
    a, b = list(map(int, input().split()))
    s, sa, sb = S[a] + S[b], S[a], S[b]
    if len(s) > 30:
        S.append(s[:10] + s[-10:])
    else:
        S.append(s)

    ai = i+N+1

    d = max(D[a], D[b]) + 1
    for dv in range(d, 10):
        if len(sa) + len(sb) < dv:
            break
        A[ai][dv] = A[a][dv] | A[b][dv] | {int(v, 2) for v in {sa[-i:] + sb[:dv-i] for i in range(1, dv+1)} if len(v) == dv}

    ans = d-1
    for dv in range(d, 10):
        if any(v not in A[ai][dv] for v in range(2**dv)):
            break
        ans = dv
    print(ans)
    D[ai] = ans





# print(time.time() - t0)




