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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  char op;
  int num;
  int t[15];
  for (int i = 1; i <= 10; ++i) t[i] = 2;
  for (int i = 0; i < n; ++i) {
    cin >> op >> num;
    for (int i = 1; i <= 10; ++i) {
      int rnum = num % 2;
      num /= 2;
      if (op == '|') {
        if (rnum == 1) t[i] = 1;
      }
      if (op == '&') {
        if (rnum == 0) t[i] = 0;
      }
      if (op == '^') {
        if (rnum == 1) {
          if (t[i] == 0)
            t[i] = 1;
          else if (t[i] == 1)
            t[i] = 0;
          else if (t[i] == 2)
            t[i] = 3;
          else if (t[i] == 3)
            t[i] = 2;
        }
      }
    }
  }
  int lAnd = 1023;
  int lOr = 0;
  int lXor = 0;
  int p2 = 1;
  for (int i = 1; i <= 10; ++i) {
    if (t[i] == 0) lAnd -= p2;
    if (t[i] == 1) lOr += p2;
    if (t[i] == 3) lXor += p2;
    p2 *= 2;
  }
  cout << 3 << "\n";
  cout << "& " << lAnd << "\n";
  cout << "| " << lOr << "\n";
  cout << "^ " << lXor << "\n";
}
