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;
int nth_bit(int64_t num, int n) { return (num >> n) & 1; }
int main() {
  int N;
  cin >> N;
  vector<char> C(N);
  vector<int> A(N);
  for (int i = 0; i < N; i++) cin >> C[i] >> A[i];
  vector<int> num(2);
  for (int k = 0; k < 2; k++) {
    int x = k * 1023;
    for (int i = 0; i < N; i++) {
      if (C[i] == '|') {
        x |= A[i];
      } else if (C[i] == '&') {
        x &= A[i];
      } else {
        x ^= A[i];
      }
    }
    num[k] = x;
  }
  int And = 1023, Xor = 0, Or = 0;
  for (int k = 0; k < 10; k++) {
    int b0 = nth_bit(num[0], k);
    int b1 = nth_bit(num[1], k);
    if (b0 && b1) {
      Or |= 1 << k;
    } else if (!b0 && !b1) {
      And ^= 1 << k;
    } else if (b0 && !b1) {
      Xor |= 1 << k;
    }
  }
  cout << 3 << endl;
  cout << "& " << And << endl;
  cout << "^ " << Xor << endl;
  cout << "| " << Or << endl;
  return 0;
}