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.
from math import log
n = int(input())
p = [bin(p)[2:] for p in range(0,512)]
def mset(s):
    ss = set()
    for k in range(0,10):
        for pi in range(0,2 ** k):
            cs = p[pi]
            cs = (k - len(cs)) * "0" + cs
            if cs in s:
                ss.add(cs)
    return ss
def q(s):
    for k in range(0,10):
        for pi in range(0,2 ** k):
            cs = p[pi]
            cs = (k - len(cs)) * "0" + cs
            if not cs in s:
                return k - 1
s = [[v[:9], v[-9:], mset(v)] for v in [input() for i in range(n)]]
for qa, qb in [[int(v) - 1 for v in input().split()] for i in range(int(input()))]:
    v = [s[qa][0], s[qb][1], mset(s[qa][1] + s[qb][0]) | s[qa][2] | s[qb][2]]
    if len(v[0]) < 9:
        v[0] = (v[0] + s[qb][0])[:9]
    if len(v[1]) < 9:
        v[1] = (s[qa][1] + s[qb][1])[-9:]
    s += [v]
    print(max(q(v[2]),0))

