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 = 5e5 + 50;
struct Operator {
  char ch;
  int x;
} p[maxn];
int n;
int cal(int x) {
  for (int i = 0; i < n; i++) {
    if (p[i].ch == '|')
      x |= p[i].x;
    else if (p[i].ch == '&')
      x &= p[i].x;
    else if (p[i].ch == '^')
      x ^= p[i].x;
  }
  return x;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].ch >> p[i].x;
  int an = 0;
  int xo = 0;
  int orr = 0;
  int anda = 0;
  int ora = 0;
  int xora = 0;
  int pos = cal(0);
  for (int i = 0; i < 10; i++) {
    int cur = cal((1 << i));
    if (((cur & (1 << i)) == 0) && ((pos & (1 << i)) == 0))
      continue;
    else if ((cur & (1 << i)) && (pos & (1 << i))) {
      an += (1 << i);
      orr += (1 << i);
    } else if (((cur & (1 << i)) == 0) && (pos & (1 << i))) {
      an += (1 << i);
      xo += (1 << i);
    } else
      an += (1 << i);
  }
  printf("3\n");
  printf("| %d\n", orr);
  printf("& %d\n", an);
  printf("^ %d\n", xo);
  return 0;
}