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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
const long long mod = 5000000;
const long long inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.000000000000000000001;
using namespace std;
long long T[12], T2[12];
long long funkcja1(long long x) { return (((x | 3) ^ 2) | 1); }
long long funkcja2(long long x) { return (((x | 1) & 1021) ^ 0); }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a = 0, b = 0, c = 0;
  long long n;
  cin >> n;
  for (long long i = 0; i < 10; i++) T[i] = 1;
  for (long long i = 1; i <= n; i++) {
    char c;
    long long x;
    cin >> c >> x;
    for (long long j = 0; j < 10; j++) {
      long long b = 0;
      if ((1 << j) & x) b = 1;
      if (c == '|')
        T[j] |= b, T2[j] |= b;
      else if (c == '&')
        T[j] &= b, T2[j] &= b;
      else
        T[j] ^= b, T2[j] ^= b;
    }
  }
  b = 1023;
  for (long long i = 0; i < 10; i++) {
    if (T[i] == 0) {
      if (T2[i] == 1)
        c += (1 << i);
      else
        b &= (1023 ^ (1 << i));
    } else if (T[i] == 1 && T2[i] == 1)
      a += (1 << i);
  }
  cout << "3\n";
  cout << "| " << a << "\n";
  cout << "& " << b << "\n";
  cout << "^ " << c << "\n";
  return 0;
}