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;
bool f[10][2];
int N;
int AND, XOR;
inline bool POS(int a, int pos) { return (((1 << pos) & a) != 0); }
int main() {
  cin >> N;
  char tipo;
  int c;
  for (int i = 0; i < 10; i++) f[i][0] = 0, f[i][1] = 1;
  while (N--) {
    cin >> tipo >> c;
    if (tipo == '&')
      for (int i = 0; i < 10; i++) {
        f[i][0] = (f[i][0] & POS(c, i));
        f[i][1] = (f[i][1] & POS(c, i));
      }
    else if (tipo == '|')
      for (int i = 0; i < 10; i++) {
        f[i][0] = (f[i][0] | POS(c, i));
        f[i][1] = (f[i][1] | POS(c, i));
      }
    else
      for (int i = 0; i < 10; i++) {
        f[i][0] = (f[i][0] ^ POS(c, i));
        f[i][1] = (f[i][1] ^ POS(c, i));
      }
  }
  for (int i = 0; i < 10; i++) {
    if (f[i][0] == 0 && f[i][1] == 0)
      ;
    if (f[i][0] == 0 && f[i][1] == 1) AND += (1 << i);
    if (f[i][0] == 1 && f[i][1] == 0) AND += (1 << i), XOR += (1 << i);
    if (f[i][0] == 1 && f[i][1] == 1) XOR += (1 << i);
  }
  cout << endl << "2" << endl << "& " << AND << endl << "^ " << XOR << endl;
  return 0;
}