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;
int flip[11];
int force[11];
int n;
void parse(int x, char c) {
  for (int i = 0; i < 10; i++) {
    int digit = x % 2;
    x /= 2;
    if (c == '|' && digit == 1) {
      force[i] = 1;
      flip[i] = 0;
    } else if (c == '&' && digit == 0) {
      force[i] = 0;
      flip[i] = 0;
    } else if (c == '^' && digit == 1) {
      flip[i] = flip[i] ^ 1;
    }
  }
}
int main(int argc, char const *argv[]) {
  memset(flip, 0, sizeof(flip));
  memset(force, -1, sizeof(force));
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    char c;
    cin >> c >> x;
    parse(x, c);
  }
  int andx = 0;
  int orx = 0;
  int xorx = 0;
  for (int i = 0; i < 10; i++) {
    int bit = 1 << i;
    if (force[i] != 0) {
      andx = andx | bit;
    }
    if (force[i] == 1) {
      orx = orx | bit;
    }
    if (flip[i] == 1) {
      xorx = xorx | bit;
    }
  }
  int k = 0;
  if (andx != 1023) k++;
  if (orx != 0) k++;
  if (xorx != 0) k++;
  cout << k << endl;
  if (andx != 1023) {
    cout << "& " << andx << endl;
  }
  if (orx != 0) {
    cout << "| " << orx << endl;
  }
  if (xorx != 0) {
    cout << "^ " << xorx << endl;
  }
  return 0;
}