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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long off = 0, on = (1LL << 10) - 1;
  for (int i = 0; i < n; i++) {
    string op;
    long long x;
    cin >> op >> x;
    if (op == "|") {
      off |= x;
      on |= x;
    } else if (op == "^") {
      off ^= x;
      on ^= x;
    } else {
      off &= x;
      on &= x;
    }
  }
  long long alwaysoff = 0, alwayson = 0, inverted = 0;
  for (int i = 0; i < 10; i++) {
    long long bit = (1LL << i);
    if ((off & bit) == 0 && (on & bit) == 0) {
      alwaysoff |= bit;
    } else if ((off & bit) != 0 && (on & bit) != 0) {
      alwayson |= bit;
    } else if ((off & bit) != 0 && (on & bit) == 0) {
      inverted |= bit;
    } else {
    }
  }
  cout << 3 << endl;
  cout << "& " << ((~alwaysoff) & ((1LL << 10) - 1)) << endl;
  cout << "| " << alwayson << endl;
  cout << "^ " << inverted << endl;
  return 0;
}
