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 N = 1e6 + 7;
string x, t;
char s;
int n, a, ors, ands, xors, c, st = -1;
int main() {
  for (int i = 0; i < 10; i++) x += 'x';
  cin >> n;
  for (int i = 0; i < n; i++) {
    int q = 10;
    t = "";
    cin >> s >> a;
    while (q--) {
      if (a & 1)
        t += '1';
      else
        t += '0';
      a = a >> 1;
    }
    for (int j = 0; j < t.length(); j++) {
      if (s == '&' && t[j] == '0') x[j] = '0';
      if (s == '|' && t[j] == '1') x[j] = '1';
      if (s == '^' && t[j] == '1') {
        if (x[j] == 'x')
          x[j] = 'y';
        else if (x[j] == 'y')
          x[j] = 'x';
        else
          x[j] = x[j] == '1' ? '0' : '1';
      }
    }
  }
  for (int i = 0; i < 10; i++)
    if (x[i] == '1') ors += (1 << i);
  for (int i = 0; i < 10; i++)
    if (x[i] != '0') ands += (1 << i);
  for (int i = 0; i < 10; i++)
    if (x[i] == 'y') xors += (1 << i);
  cout << 3 << endl;
  cout << "^ " << xors << endl;
  cout << "| " << ors << endl;
  cout << "& " << ands;
  return 0;
}