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;
using ll = long long;
using ull = unsigned long long;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 1e6 + 8;
int N, M, K, Q;
char op[MAXN];
int x[MAXN];
int ans[3];
void preprocess() {}
void solution() {
  int zeros = 0, ones = 1023;
  for (int i = 0; i < N; ++i) {
    switch (op[i]) {
      case '|':
        zeros |= x[i], ones |= x[i];
        break;
      case '&':
        zeros &= x[i], ones &= x[i];
        break;
      case '^':
        zeros ^= x[i], ones ^= x[i];
        break;
    }
  }
  ans[0] = ones | zeros;
  ans[1] = ones & zeros;
  ans[2] = (ones ^ zeros) & zeros;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; ++i) cin >> op[i] >> x[i];
  solution();
  cout << "3\n";
  cout << "& " << ans[0] << '\n';
  cout << "| " << ans[1] << '\n';
  cout << "^ " << ans[2] << '\n';
  return 0;
}