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 program(int x, vector<char> &op, vector<int> &a) {
  int n = op.size();
  for (int i = 0; i < n; i++) {
    if (op[i] == '&')
      x &= a[i];
    else if (op[i] == '^')
      x ^= a[i];
    else
      x |= a[i];
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  int n;
  cin >> n;
  vector<char> op(n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> op[i] >> a[i];
  int andnum = 1023;
  int ornum = 0, xornum = 0;
  for (int i = 0; i < 10; i++) {
    int x = 1 << i;
    int y1 = program(x, op, a) & x;
    int y0 = program(0, op, a) & x;
    if (y0 == 0 and y1 == 0) andnum &= (1023 - x);
    if (y0 != 0 and y1 != 0) ornum |= x;
    if (y0 != 0 and y1 == 0) xornum |= x;
  }
  cout << 3 << endl;
  cout << "& " << andnum << endl;
  cout << "^ " << xornum << endl;
  cout << "| " << ornum << endl;
  return 0;
}
