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;
struct cmd {
  char op;
  int val;
  int calc(int a) {
    switch (op) {
      case '|':
        return a | val;
      case '&':
        return a & val;
      case '^':
        return a ^ val;
    }
  }
};
int execute(int val, vector<cmd> &prog) {
  for (auto &x : prog) {
    val = x.calc(val);
  }
  return val;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int N;
  cin >> N;
  vector<cmd> prog;
  prog.reserve(N);
  for (int i = 0; i < N; i++) {
    char c;
    int v;
    cin >> c >> v;
    prog.push_back({c, v});
  }
  int a = execute(0, prog);
  int b = execute(1023, prog);
  cout << 3 << '\n';
  cout << "^ " << (a | (b ^ 1023)) << '\n';
  cout << "& " << ~(~a & ~b) << '\n';
  cout << "| " << (a & b) << '\n';
}