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() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int n, x;
  cin >> n;
  string s;
  vector<int> z(10);
  vector<int> o(10);
  vector<int> r(10);
  for (int i = 0; i < n; ++i) {
    cin >> s >> x;
    for (int b = 0; b < 10; ++b) {
      bool t = x & (1 << b);
      if (s[0] == '|' && t) {
        o[b] = 1;
        z[b] = 0;
      }
      if (s[0] == '&' && !t) {
        z[b] = 1;
        o[b] = 0;
      }
      if (s[0] == '^' && t) {
        if (z[b]) {
          z[b] = 0;
          o[b] = 1;
        } else if (o[b]) {
          o[b] = 0;
          z[b] = 1;
        }
        r[b] ^= 1;
      }
    }
  }
  cout << 3 << endl;
  x = 0;
  for (int i = 0; i < 10; ++i)
    if (o[i] + z[i] == 0 && r[i]) x ^= 1 << i;
  cout << "^ " << x << endl;
  x = 0;
  for (int i = 0; i < 10; ++i)
    if (o[i]) x ^= 1 << i;
  cout << "| " << x << endl;
  x = 1023;
  for (int i = 0; i < 10; ++i)
    if (z[i]) x ^= 1 << i;
  cout << "& " << x << endl;
  return 0;
}