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;
const int nax = 5e5 + 11;
int n;
int x[nax];
char c[nax];
map<char, int> typ;
int go(int k, int w) {
  for (int i = 1; i <= n; ++i) {
    if (typ[c[i]] == 0) w |= ((x[i] >> k) & 1);
    if (typ[c[i]] == 1) w ^= ((x[i] >> k) & 1);
    if (typ[c[i]] == 2) w &= ((x[i] >> k) & 1);
  }
  return w;
}
int mask[3];
int main() {
  typ['|'] = 0;
  typ['^'] = 1;
  typ['&'] = 2;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf(" %c %d", &c[i], &x[i]);
  }
  for (int i = 0; i < 10; ++i) {
    int a = go(i, 0);
    int b = go(i, 1);
    if (a == 0 && b == 1) mask[2] |= (1 << i);
    if (a == 1 && b == 0) {
      mask[1] |= (1 << i);
      mask[2] |= (1 << i);
    }
    if (a == 1 && b == 1) {
      mask[0] |= (1 << i);
      mask[2] |= (1 << i);
    }
  }
  printf("3\n");
  printf("| %d\n", mask[0]);
  printf("^ %d\n", mask[1]);
  printf("& %d\n", mask[2]);
  return 0;
}
