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.
#include <bits/stdc++.h>
using namespace std;
const int N = 1048576, M = 10;
const int SAME = 0;
const int DIFF = 1;
const int ONE = 2;
const int ZERO = 3;
int n, num, a[N];
char op;
inline bool is_set(int x, int b) { return x >> b & 1; }
int main() {
  cin >> n;
  while (n--) {
    scanf(" %c %d", &op, &num);
    if (op == '|') {
      for (int i = 0; i < M; ++i)
        if (is_set(num, i)) a[i] = ONE;
    } else if (op == '&') {
      for (int i = 0; i < M; ++i)
        if (!is_set(num, i)) a[i] = ZERO;
    } else {
      for (int i = 0; i < M; ++i)
        if (is_set(num, i)) a[i] ^= 1;
    }
  }
  int o = 0;
  for (int i = 0; i < M; ++i)
    if (a[i] == ONE) o |= 1 << i;
  int ad = (1 << M) - 1;
  for (int i = 0; i < M; ++i)
    if (a[i] == ZERO) ad ^= 1 << i;
  int x = 0;
  for (int i = 0; i < M; ++i)
    if (a[i] == DIFF) x |= 1 << i;
  cout << 3 << endl;
  cout << "| " << o << endl;
  cout << "& " << ad << endl;
  cout << "^ " << x << endl;
  return 0;
}
