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;
const int N = 2e5 + 11;
int d[10];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int ii = 1; ii <= n; ii++) {
    char ch;
    int l;
    cin >> ch >> l;
    if (ch == '|') {
      for (int i = 0; i <= 9; i++)
        if (l & (1 << i)) d[i] = 1;
    }
    if (ch == '^') {
      for (int i = 0; i <= 9; i++)
        if (l & (1 << i)) {
          if (d[i] == -1)
            d[i] = 1;
          else if (d[i] == 1)
            d[i] = -1;
          else if (d[i] == 0)
            d[i] = 2;
          else
            d[i] = 0;
        }
    }
    if (ch == '&') {
      for (int i = 0; i <= 9; i++)
        if (l & (1 << i)) {
        } else
          d[i] = -1;
    }
  }
  int d1 = 0, d2 = 0, d3 = 0;
  for (int i = 0; i <= 9; i++) {
    if (d[i] == 1)
      d1 ^= (1 << i);
    else if (d[i] == 2)
      d2 ^= (1 << i);
    if (d[i] != -1) d3 ^= (1 << i);
  }
  vector<char> v1;
  vector<int> v2;
  {
    v1.push_back('&');
    v2.push_back(d3);
  }
  {
    v1.push_back('|');
    v2.push_back(d1);
  }
  {
    v1.push_back('^');
    v2.push_back(d2);
  }
  cout << v1.size() << endl;
  for (int i = 0; i < v1.size(); i++) cout << v1[i] << " " << v2[i] << endl;
}
