Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, x1 = 1023, x0, ans[3] = {0, 0, 1023}, t;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int a;
    char b;
    cin >> b >> a;
    if (b == '|') {
      x1 |= a;
      x0 |= a;
    } else if (b == '&') {
      x0 &= a;
      x1 &= a;
    } else {
      x0 ^= a;
      x1 ^= a;
    }
  }
  for (int i = 0; i < 10; ++i) {
    int a = (x0 & (1 << i)), b = (x1 & (1 << i));
    if (a) a = 1;
    if (b) b = 1;
    if (a == b && a == 0)
      ans[2] ^= (1 << i);
    else if (a == b && a == 1)
      ans[1] ^= (1 << i);
    else if (a != b && a == 1)
      ans[0] ^= (1 << i);
  }
  cout << "3\n";
  cout << "^ " << ans[0] << endl
       << "| " << ans[1] << endl
       << "& " << ans[2] << endl;
  return 0;
}