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 MOD = 1000000007;
long double PI = 4 * atan(1);
int n, fct[500001], x1[10], x2[10], x3[10], x4[10], cur;
char sym[500001];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> sym[i] >> fct[i];
  }
  for (int i = 0; i < 10; i++) {
    x1[i] = 0, x2[i] = 1, x3[i] = 1, x4[i] = 0;
    int res1 = 0, res2 = 1023;
    for (int j = 0; j < n; j++) {
      if (sym[j] == '|') {
        res1 = res1 | fct[j];
        res2 = res2 | fct[j];
      } else if (sym[j] == '&') {
        res1 = res1 & fct[j];
        res2 = res2 & fct[j];
      } else if (sym[j] == '^') {
        res1 = res1 ^ fct[j];
        res2 = res2 ^ fct[j];
      }
    }
    if (res1 & (1 << i)) {
      res1 = 1;
    } else {
      res1 = 0;
    }
    if (res2 & (1 << i)) {
      res2 = 1;
    } else {
      res2 = 0;
    }
    if (res1 == 1 && res2 == 1)
      x1[i] = 1;
    else if (res1 == 1 && res2 == 0)
      x4[i] = 1;
    else if (res1 == 0 && res2 == 0)
      x3[i] = 0;
    else {
      x2[i] = 1;
    }
  }
  cout << 4 << "\n";
  cout << '|' << " ";
  cur = 0;
  for (int i = 0; i < 10; i++) {
    cur += (pow(2, i) * x1[i]);
  }
  cout << cur << "\n";
  cout << '&' << " ";
  cur = 0;
  for (int i = 0; i < 10; i++) {
    cur += (pow(2, i) * x2[i]);
  }
  cout << cur << "\n";
  cout << '&' << " ";
  cur = 0;
  for (int i = 0; i < 10; i++) {
    cur += (pow(2, i) * x3[i]);
  }
  cout << cur << "\n";
  cout << '^' << " ";
  cur = 0;
  for (int i = 0; i < 10; i++) {
    cur += (pow(2, i) * x4[i]);
  }
  cout << cur << "\n";
  return 0;
}