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.
#include <bits/stdc++.h>
using namespace std;
int n, x, y, k, ans, a, b, c;
char s[5];
int main() {
  scanf("%d", &n);
  x = 0;
  y = 1023;
  while (n--) {
    scanf("%s", s);
    scanf("%d", &k);
    if (s[0] == '|') x |= k, y |= k;
    if (s[0] == '^') x ^= k, y ^= k;
    if (s[0] == '&') x &= k, y &= k;
  }
  a = x & y;
  b = 1023 - ((1023 - x) & (1023 - y));
  c = (1023 - y) & x;
  ans = (a > 0) + (b < 1023) + (c > 0);
  printf("%d\n", ans);
  if (a > 0) printf("| %d\n", a);
  if (b < 1023) printf("& %d\n", b);
  if (c > 0) printf("^ %d\n", c);
  return 0;
}
