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 get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
int v1[11], v0[11];
int main() {
  int n = get();
  int now = 1023;
  for (int j = 0; j <= 9; j++) v1[j] = 1, v0[j] = 0;
  for (int i = 1; i <= n; i++) {
    char ch;
    while (ch = getchar(), ch != '&' && ch != '|' && ch != '^')
      ;
    int x = get();
    if (ch == '&') {
      for (int j = 0; j <= 9; j++)
        if ((x & (1 << j)) == 0) v1[j] = v0[j] = 0;
    }
    if (ch == '|') {
      for (int j = 0; j <= 9; j++)
        if ((x & (1 << j)) > 0) v1[j] = 0, v0[j] = 1;
    }
    if (ch == '^') {
      for (int j = 0; j <= 9; j++)
        if ((x & (1 << j)) > 0) v0[j] ^= 1;
    }
  }
  printf("2\n");
  int tmp = 0;
  for (int i = 0; i <= 9; i++)
    if (v1[i]) tmp = tmp + (1 << i);
  printf("& %d\n", tmp);
  tmp = 0;
  for (int i = 0; i <= 9; i++)
    if (v0[i]) tmp += 1 << i;
  printf("^ %d\n", tmp);
  return 0;
}