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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int a = 0, b = 1023, c, n;
  char op;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> op >> c;
    switch (op) {
      case '|':
        a |= c;
        b |= c;
        break;
      case '&':
        a &= c;
        b &= c;
        break;
      case '^':
        a ^= c;
        b ^= c;
        break;
    }
  }
  int r = 0, nd = 1023, xr = 0;
  for (int i = 0; i < 10; ++i) {
    int p = (a >> i) & 1;
    int q = (b >> i) & 1;
    if (!p && !q) {
      nd ^= (1 << i);
    } else if (p && !q) {
      xr |= (1 << i);
    } else if (p && q) {
      r |= (1 << i);
    }
  }
  int s = 3;
  if (nd == 1023) s--;
  if (xr == 0) s--;
  if (r == 0) s--;
  cout << s << endl;
  if (nd != 1023) cout << "& " << nd << endl;
  if (xr != 0) cout << "^ " << xr << endl;
  if (r != 0) cout << "| " << r << endl;
  return 0;
}