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 n;
int bitDefined[15];
char c[500500];
int x[500500];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 0; i < 15; i++) bitDefined[i] = -1;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i] >> x[i];
  for (int i = n; i >= 1; i--) {
    if (c[i] == '|') {
      for (int b = 0; b < 10; b++) {
        if (!((1 << b) & x[i])) continue;
        if (bitDefined[b] == 0) continue;
        if (bitDefined[b] == -1) bitDefined[b] = 1;
        if (bitDefined[b] == -2) bitDefined[b] = 0;
      }
    } else if (c[i] == '&') {
      for (int b = 0; b < 10; b++) {
        if ((1 << b) & x[i]) continue;
        if (bitDefined[b] == 1) continue;
        if (bitDefined[b] == -1) bitDefined[b] = 0;
        if (bitDefined[b] == -2) bitDefined[b] = 1;
      }
    } else {
      for (int b = 0; b < 10; b++) {
        if (!((1 << b) & x[i])) continue;
        if (bitDefined[b] >= 0) continue;
        if (bitDefined[b] == -1)
          bitDefined[b]--;
        else
          bitDefined[b]++;
      }
    }
  }
  int toAnd = 0;
  int toOr = 0;
  int toXor = 0;
  for (int b = 0; b < 10; b++) {
    if (bitDefined[b] == 1)
      toOr |= (1 << b);
    else if (bitDefined[b] == 0)
      toAnd |= (1 << b);
    else if (bitDefined[b] == -2)
      toXor |= (1 << b);
  }
  toAnd = ~toAnd;
  toAnd &= (1 << 10) - 1;
  cout << 3 << endl;
  cout << "^ " << toXor << endl;
  cout << "& " << toAnd << endl;
  cout << "| " << toOr << endl;
  return 0;
}