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 long long oo = (long long)1e18;
const int mod = (int)1e9 + 7;
const double PI = acos((long long)-1.0);
class TaskA {
 public:
  void solve(std::istream& in, std::ostream& out) {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, a[15];
    fill(a, a + 11, 3);
    in >> n;
    for (int i = 0; i < n; ++i) {
      char c;
      int x;
      in >> c >> x;
      if (c == '^') {
        for (int j = 0; j < 10; ++j) {
          if ((1 << j) & x) {
            if (a[j] == 0)
              a[j] = 1;
            else if (a[j] == 1)
              a[j] = 0;
            else if (a[j] == 2)
              a[j] = 3;
            else if (a[j] == 3)
              a[j] = 2;
          }
        }
      }
      if (c == '&') {
        for (int j = 0; j < 10; ++j) {
          if (((1 << j) & x) == 0) a[j] = 0;
        }
      }
      if (c == '|') {
        for (int j = 0; j < 10; ++j) {
          if ((1 << j) & x) a[j] = 1;
        }
      }
    }
    int x = 0, y = 1023, z = 0;
    for (int i = 0; i < 10; ++i) {
      if (a[i] == 0) y = y - (1 << i);
      if (a[i] == 1) z = z + (1 << i);
      if (a[i] == 2) x = x + (1 << i);
    }
    out << 3 << "\n";
    out << "& " << y << "\n";
    out << "^ " << x << "\n";
    out << "| " << z << "\n";
  };
};
int main() {
  TaskA solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}