Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well.

In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned.

Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023.

Input

The first line contains an integer n (1 ≤ n ≤ 5·105) — the number of lines.

Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023.

Output

Output an integer k (0 ≤ k ≤ 5) — the length of your program.

Next k lines must contain commands in the same format as in the input.

Examples

Input

3
| 3
^ 2
| 1


Output

2
| 3
^ 2


Input

3
&amp; 1
&amp; 3
&amp; 5


Output

1
&amp; 1


Input

3
^ 1
^ 2
^ 3


Output

0

Note

You can read about bitwise operations in <https://en.wikipedia.org/wiki/Bitwise_operation>.

Second sample:

Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs.
#include <bits/stdc++.h>
using std::vector;
size_t operation(size_t x, size_t y, char c) {
  if (c == '|') {
    return x | y;
  }
  if (c == '^') {
    return x ^ y;
  }
  if (c == '&') {
    return x & y;
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  size_t n;
  std::cin >> n;
  vector<size_t> bits0(10, 0), bits1(10, 1);
  for (size_t i = 0; i < n; ++i) {
    char x;
    size_t y;
    std::cin >> x >> y;
    for (size_t j = 0; j < 10; ++j) {
      size_t z = !!(y & 1 << j);
      bits0[j] = operation(bits0[j], z, x);
      bits1[j] = operation(bits1[j], z, x);
    }
  }
  size_t cnt_or = 0, cnt_xor = 0, cnt_and = (1 << 10) - 1;
  for (size_t i = 0; i < 10; ++i) {
    size_t x = 1 << i;
    if (bits0[i] == 1 && bits1[i] == 1) cnt_or |= x;
    if (bits0[i] == 0 && bits1[i] == 0) cnt_and &= ~x;
    if (bits0[i] == 1 && bits1[i] == 0) cnt_xor |= x;
  }
  std::cout << 3 << '\n'
            << "| " << cnt_or << '\n'
            << "^ " << cnt_xor << '\n'
            << "& " << cnt_and;
  return 0;
}
