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() {
  int n;
  cin >> n;
  int a, b;
  a = 1023;
  b = 0;
  int l = 10;
  for (int i = 0; i < n; ++i) {
    char c;
    int num;
    cin >> c >> num;
    if (c == '&') {
      a &= num;
      b &= num;
    } else if (c == '|') {
      a |= num;
      b |= num;
    } else {
      a ^= num;
      b ^= num;
    }
  }
  vector<int> v[4];
  for (int i = 0; i < 10; ++i) {
    int _a, _b;
    _a = a & 1;
    _b = b & 1;
    a >>= 1;
    b >>= 1;
    if (_a == 1 && _b == 0) {
      v[0].push_back(i);
    } else if (_a == 1 && _b == 1) {
      v[1].push_back(i);
    } else if (_a == 0 && _b == 0) {
      v[2].push_back(i);
    } else
      v[3].push_back(i);
  }
  int or_val = 0;
  for (int i = 0; i < v[1].size(); ++i) {
    or_val |= (1 << (v[1][i]));
  }
  int and_val = 1023;
  for (int i = 0; i < v[2].size(); ++i) {
    and_val -= (1 << v[2][i]);
  }
  int xor_val = 0;
  for (int i = 0; i < v[3].size(); ++i) {
    xor_val |= (1 << v[3][i]);
  }
  cout << 3 << endl;
  cout << "| " << or_val << endl;
  cout << "& " << and_val << endl;
  cout << "^ " << xor_val << endl;
  return 0;
}