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;
long long int status(long long int a, long long int b, long long int i) {
  a >>= i;
  b >>= i;
  a = a % 2;
  b = b % 2;
  if (a == 0)
    if (b == 1)
      return 1;
    else
      return 4;
  if (b == 1) return 3;
  return 2;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a = 0, b = 1023;
  for (long long int i = 0; i < n; ++i) {
    char c;
    long long int x;
    cin >> c >> x;
    if (c == '^') {
      a ^= x;
      b ^= x;
    } else if (c == '|') {
      a |= x;
      b |= x;
    } else {
      a &= x;
      b &= x;
    }
  }
  long long int x = 0, o = 0, ad = 0;
  long long int xr = 0, ore = 0, adr = 0;
  for (long long int i = 0; i < 10; ++i) {
    long long int c = status(a, b, i);
    if (c == 1) {
      ad += pow(2, i);
    } else if (c == 2) {
      xr = 1;
      x += pow(2, i);
      ad += pow(2, i);
    } else if (c == 3) {
      ore = 1;
      o += pow(2, i);
      ad += pow(2, i);
    } else {
      adr = 1;
    }
  }
  cout << ore + xr + adr << endl;
  if (ore == 1) cout << "| " << o << endl;
  if (xr == 1) cout << "^ " << x << endl;
  if (adr == 1) cout << "& " << ad << endl;
}
int main() {
  solve();
  return 0;
}