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;
char m[12];
void print() {
  int i = 0;
  for (i = 0; i < 11; i++) cout << m[i] << " ";
  cout << "\n";
}
int main() {
  int n;
  cin >> n;
  int i;
  for (i = 0; i < 12; i++) m[i] = 'a';
  for (i = 0; i < n; i++) {
    string c;
    int x;
    cin >> c >> x;
    int j;
    if (c == "|") {
      for (j = 0; j < 11; j++) {
        if ((1 << j) & x) {
          m[j] = 'n';
        }
      }
    } else if (c == "^") {
      for (j = 0; j < 11; j++) {
        if ((1 << j) & x) {
          if (m[j] == 'n')
            m[j] = 'o';
          else if (m[j] == 'o')
            m[j] = 'n';
          else if (m[j] == 'a')
            m[j] = 'f';
          else
            m[j] = 'a';
        }
      }
    } else {
      for (j = 0; j < 11; j++)
        if (((1 << j) & x) == 0) m[j] = 'o';
    }
  }
  int o = 0, a = 0;
  int x = 0;
  for (int j = 0; j < 10; j++) {
    if (m[j] == 'f') {
      x = x | (1 << j);
    } else if (m[j] == 'n') {
      o = o | (1 << j);
    }
    if (m[j] != 'o') a = a | (1 << j);
  }
  cout << 3 << "\n";
  cout << "& " << a << "\n";
  cout << "| " << o << "\n";
  cout << "^ " << x << "\n";
  return 0;
}