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 li = long long;
using ld = long double;
void solve(bool read) {
  int n;
  cin >> n;
  const int max_bits = 10;
  const int X = 0, FLIP_X = 1, ZERO = 2, ONE = 3;
  vector<int> type(max_bits);
  for (int i = 0; i < n; i++) {
    string cmd;
    int num;
    cin >> cmd >> num;
    if (cmd == "^") {
      for (int i = 0; i < max_bits; i++)
        if (num & (1 << i)) type[i] ^= 1;
    } else if (cmd == "&") {
      for (int i = 0; i < max_bits; i++)
        if (!(num & (1 << i))) type[i] = ZERO;
    } else if (cmd == "|") {
      for (int i = 0; i < max_bits; i++)
        if (num & (1 << i)) type[i] = ONE;
    }
  }
  int and_mask = 0, or_mask = 0, xor_mask = 0;
  for (int i = 0; i < max_bits; i++) {
    if (type[i] != ZERO) and_mask |= (1 << i);
    if (type[i] == ONE) or_mask |= (1 << i);
    if (type[i] == FLIP_X) xor_mask |= (1 << i);
  }
  cout << "3\n";
  cout << "^ " << xor_mask << "\n";
  cout << "& " << and_mask << "\n";
  cout << "| " << or_mask << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int nt = 1;
  for (int i = 0; i < nt; i++) solve(true);
  return 0;
}