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 N = 5e5 + 10;
const int inf = 1e18 + 5;
const int mod = 1e9 + 7;
int n;
int a[10][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int j = 0; j < 10; ++j) a[j][1] = 1;
  for (int i = 0; i < n; ++i) {
    char c;
    int x;
    cin >> c >> x;
    for (int j = 0; j < 10; ++j) {
      bool bit = x & (1 << j);
      if (c == '|') {
        a[j][0] |= bit;
        a[j][1] |= bit;
      } else if (c == '&') {
        a[j][0] &= bit;
        a[j][1] &= bit;
      } else {
        a[j][0] ^= bit;
        a[j][1] ^= bit;
      }
    }
  }
  int flag1 = 0;
  int flag2 = 0;
  int flag3 = 0;
  int x1 = 0;
  int x2 = 0;
  int x3 = 0;
  for (int j = 0; j < 10; ++j) {
    if (a[j][0] == 0 && a[j][1] == 1) {
    } else if (a[j][0] == 1 && a[j][1] == 1) {
      flag1 = 1;
      x1 |= (1 << j);
    } else if (a[j][0] == 0 && a[j][1] == 0) {
      flag2 = 1;
      x2 |= (1 << j);
    } else {
      flag3 = 1;
      x3 |= (1 << j);
    }
  }
  cout << flag1 + flag2 + flag3 << endl;
  if (flag1) {
    cout << "| " << x1 << endl;
  }
  if (flag2) {
    cout << "& " << (x2 ^ (1023)) << endl;
  }
  if (flag3) {
    cout << "^ " << x3 << endl;
  }
  return 0;
}