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;
const int maxN = 5e4;
int mask[10];
void applyOr(int key) {
  for (int i = 0; i < 10; i++) {
    if (mask[i] < 2) {
      mask[i] |= ((1 << i) & key) ? 1 : 0;
    } else {
      if ((1 << i) & key) mask[i] = 1;
    }
  }
}
void applyAnd(int key) {
  for (int i = 0; i < 10; i++) {
    if (mask[i] < 2) {
      mask[i] &= ((1 << i) & key) ? 1 : 0;
    } else {
      if (((1 << i) & key) == 0) mask[i] = 0;
    }
  }
}
void applyXor(int key) {
  for (int i = 0; i < 10; i++) {
    mask[i] ^= ((1 << i) & key) ? 1 : 0;
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  for (int i = 0; i < 10; i++) {
    mask[i] = 2;
  }
  int n;
  char c;
  cin >> n;
  for (int i = 0, tmp; i < n; i++) {
    cin >> c >> tmp;
    switch (c) {
      case '|':
        applyOr(tmp);
        break;
      case '&':
        applyAnd(tmp);
        break;
      default:
        applyXor(tmp);
    }
  }
  int xorKey = 0, orKey = 0, andKey = 1023;
  for (int i = 0; i < 10; i++) {
    if (mask[i] == 3)
      xorKey |= 1 << i;
    else if (mask[i] == 1)
      orKey |= 1 << i;
    else if (mask[i] == 0)
      andKey ^= 1 << i;
  }
  cout << "3\n^ " << xorKey << "\n& " << andKey << "\n| " << orKey << '\n';
}