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 int MAXB = 10;
array<bool, 2> f[MAXB];
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i(0); i < MAXB; ++i) {
    f[i][0] = false;
    f[i][1] = true;
  }
  for (int i(0); i < n; ++i) {
    char op;
    int arg;
    cin >> op >> arg;
    for (int p(0); p < MAXB; ++p) {
      int bit = !!((1 << p) & arg);
      for (int b(0); b < 2; ++b) {
        if (op == '|')
          f[p][b] = f[p][b] | bit;
        else if (op == '&')
          f[p][b] = f[p][b] & bit;
        else
          f[p][b] = f[p][b] ^ bit;
      }
    }
  }
  cout << 2 << endl;
  cout << "| ";
  int arg(0);
  for (int p(0); p < MAXB; ++p)
    if (f[p][0] == f[p][1]) arg += (1 << p);
  cout << arg << endl;
  arg = 0;
  cout << "^ ";
  for (int p(0); p < MAXB; ++p)
    if (f[p][1] == 0) arg += (1 << p);
  cout << arg << endl;
}