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;
const int N = 500010;
int main() {
  int n;
  scanf("%d", &n);
  char ch;
  int x;
  int n0 = 0, n1 = (1 << 10) - 1;
  for (int i = (1); i <= (n); i++) {
    while (ch = getchar(), ch != '|' && ch != '&' && ch != '^')
      ;
    scanf("%d", &x);
    if (ch == '|')
      n0 |= x, n1 |= x;
    else if (ch == '&')
      n0 &= x, n1 &= x;
    else if (ch == '^')
      n0 ^= x, n1 ^= x;
  }
  int ans0 = 0, ans1 = 0, ans2 = 0;
  for (int i = (0); i <= (9); i++) {
    if ((n0 & (1LL << (i))) || (n1 & (1LL << (i)))) ans0 |= (1LL << (i));
    if ((n0 & (1LL << (i))) && (n1 & (1LL << (i)))) ans1 |= (1LL << (i));
    if ((n0 & (1LL << (i))) && !(n1 & (1LL << (i)))) ans2 |= (1LL << (i));
  }
  printf("3\n& %d\n| %d\n^ %d\n", ans0, ans1, ans2);
  return 0;
}