Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

  1. "+ x" — add integer x to multiset A.
  2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
  3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.



Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.

Example

Input

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11


Output

11
10
14
13

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>.
MX = 29


class Trie:
    def __init__(self):
        self.children = [None] * 2
        self.count = 0


def add(trie, x):
    for i in range(MX, -1, -1):
        b = x & (1 << i) != 0
        if trie.children[b] is None:
            trie.children[b] = Trie()
        trie = trie.children[b]
    trie.count += 1


def remove(trie, x, i=MX):
    if i == -1:
        assert trie.count > 0
        trie.count -= 1
        return trie.count == 0
    j = x & (1 << i) != 0
    if remove(trie.children[j], x, i - 1):
        trie.children[j] = None
        return trie.children[j ^ 1] is None
    else:
        return False


def max_xor(trie, x):
    res = 0
    for i in range(MX, -1, -1):
        best = x & (1 << i) == 0
        if trie.children[best] is None:
            best ^= 1
        trie = trie.children[best]
        res = 2 * res + (best ^ ((x >> i) & 1))
    return res


q = int(input())
trie = Trie()
add(trie, 0)
res = []
for _ in range(q):
    o, x = input().split()
    x = int(x)
    if o == '+':
        add(trie, x)
    elif o == '-':
        remove(trie, x)
    else:
        assert o == '?'
        res.append(max_xor(trie, x))
print('\n'.join(map(str, res)))