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 main() {
  ios::sync_with_stdio(0);
  int N;
  cin >> N;
  char c[N];
  int A[N];
  for (int i = 0; i < N; i++) {
    cin >> c[i] >> A[i];
  }
  if (N <= 5) {
    cout << N << endl;
    for (int i = 0; i < N; i++) cout << c[i] << " " << A[i] << endl;
  } else {
    string bitwise[N];
    for (int i = 0; i < N; i++) {
      int temp = A[i];
      while (temp != 0) {
        if (temp % 2 == 1)
          bitwise[i].push_back('1');
        else
          bitwise[i].push_back('0');
        temp /= 2;
      }
      while (bitwise[i].size() < 10) bitwise[i].push_back('0');
      reverse(bitwise[i].begin(), bitwise[i].end());
    }
    char for0[10], for1[10];
    int num;
    for (int i = 0; i < 10; i++) {
      num = 0;
      for (int j = 0; j < N; j++) {
        if (c[j] == '|')
          num |= (bitwise[j][i] - '0');
        else if (c[j] == '&')
          num &= (bitwise[j][i] - '0');
        else
          num ^= (bitwise[j][i] - '0');
      }
      if (num == 0)
        for0[i] = '0';
      else
        for0[i] = '1';
    }
    for (int i = 0; i < 10; i++) {
      num = 1;
      for (int j = 0; j < N; j++) {
        if (c[j] == '|')
          num |= (bitwise[j][i] - '0');
        else if (c[j] == '&')
          num &= (bitwise[j][i] - '0');
        else
          num ^= (bitwise[j][i] - '0');
      }
      if (num == 0)
        for1[i] = '0';
      else
        for1[i] = '1';
    }
    int bit1[10], bit2[10] = {}, bit3[10] = {};
    for (int i = 0; i < 10; i++) bit1[i] = 1;
    for (int i = 0; i < 10; i++) {
      if (for0[i] == '0' && for1[i] == '0')
        bit1[i] = 0;
      else if (for0[i] == '1' && for1[i] == '0')
        bit2[i] = 1;
      else if (for0[i] == '1' && for1[i] == '1')
        bit3[i] = 1;
    }
    int num1 = 0, num2 = 0, num3 = 0;
    for (int i = 0; i < 10; i++) {
      num1 += pow(2, i) * bit1[10 - i - 1];
    }
    for (int i = 0; i < 10; i++) {
      num2 += pow(2, i) * bit2[10 - i - 1];
    }
    for (int i = 0; i < 10; i++) {
      num3 += pow(2, i) * bit3[10 - i - 1];
    }
    cout << 3 << endl;
    cout << "& " << num1 << endl;
    cout << "^ " << num2 << endl;
    cout << "| " << num3 << endl;
  }
  return 0;
}