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 n, m, k;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  int bits[10], len = 10, f[4]{1, 0, 3, 2};
  for (int i = (0); i < (len); i++) bits[i] = 2;
  for (int i = (0); i < (n); i++) {
    char ch;
    int x;
    cin >> ch >> x;
    if (ch == '&') {
      for (int j = (0); j < (len); j++) {
        int t = (x & (1 << j));
        if (!t) bits[j] = 0;
      }
    }
    if (ch == '|') {
      for (int j = (0); j < (len); j++) {
        int t = (x & (1 << j));
        if (t) bits[j] = 1;
      }
    }
    if (ch == '^') {
      for (int j = (0); j < (len); j++) {
        int t = (x & (1 << j));
        if (bits[j] < 2 && t)
          bits[j] = 1 - bits[j];
        else if (bits[j] > 1 && t)
          bits[j] = 5 - bits[j];
      }
    }
  }
  int And = 1023, Or = 0, Xor = 0;
  for (int i = (0); i < (len); i++)
    if (bits[i] == 0) And -= (1 << i);
  for (int i = (0); i < (len); i++)
    if (bits[i] == 1) Or += (1 << i);
  for (int i = (0); i < (len); i++)
    if (bits[i] == 3) Xor += (1 << i);
  cout << 3 << endl;
  cout << "& " << And << endl << "| " << Or << endl << "^ " << Xor << endl;
  return 0;
}