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 MAXN = 500005;
int a[MAXN];
int opt[MAXN], dt[MAXN];
int n;
char s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("\n%c %d", &s, &dt[i]);
    if (s == '|')
      opt[i] = 0;
    else if (s == '&')
      opt[i] = 1;
    else
      opt[i] = 2;
  }
  int land = (1 << 10) - 1, lxor = 0, lor = 0;
  for (int i = 0; i < 10; i++) {
    int zero = 0, one = 1;
    for (int j = 1; j <= n; j++) {
      if (opt[j] == 0)
        zero |= ((dt[j] >> i) & 1), one |= ((dt[j] >> i) & 1);
      else if (opt[j] == 1)
        one &= ((dt[j] >> i) & 1), zero &= ((dt[j] >> i) & 1);
      else {
        one ^= ((dt[j] >> i) & 1);
        zero ^= ((dt[j] >> i) & 1);
      }
    }
    if (zero == 0 && one == 1)
      ;
    else if (zero == 1 && one == 1)
      lor |= (1 << i);
    else if (zero == 0 && one == 0)
      land ^= (1 << i);
    else
      lxor |= (1 << i);
  }
  cout << 3 << endl;
  cout << "& " << land << endl;
  cout << "| " << lor << endl;
  cout << "^ " << lxor << endl;
  return 0;
}