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;
pair<char, int> pr[500100];
int arr[11];
map<vector<int>, int> mp;
vector<int> ans;
int main() {
  int n;
  cin >> n;
  bool flag = 0;
  int an = 1023, o = 0, xo = 0;
  for (int i = 1; i <= n; ++i) {
    char c;
    int x;
    cin >> c;
    scanf("%d", &x);
    for (int j = 0; j < 10; ++j) {
      if (c == '&' && ((1 << j) & x) == 0)
        an &= ~(1 << j), o &= ~(1 << j), xo &= ~(1 << j);
      else if (c == '|' && ((1 << j) & x))
        an |= (1 << j), o |= (1 << j), xo &= ~(1 << j);
      else if (c == '^' && ((1 << j) & x))
        xo ^= (1 << j);
    }
  }
  printf("3\n& %d\n| %d\n^ %d\n", an, o, xo);
  return 0;
}