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;
int table[10][2] = {};
int n;
int main() {
  cin >> n;
  int a = 0, b = 1023;
  char op;
  int operand;
  for (int i = 0; i < n; i++) {
    op = ' ';
    while ((op = getchar()) == ' ' || op == '\n' || op == '\r')
      ;
    scanf("%d", &operand);
    switch (op) {
      case '|':
        a |= operand;
        b |= operand;
        break;
      case '^':
        a ^= operand;
        b ^= operand;
        break;
      case '&':
        a &= operand;
        b &= operand;
        break;
      default:
        cout << "Illegal input" << endl;
        break;
    }
  }
  for (int i = 0; i < 10; i++) {
    table[i][0] = (a & 1);
    table[i][1] = (b & 1);
    a >>= 1;
    b >>= 1;
  }
  int ando = 1023, oro = 0, xoro = 0, curbit = 1;
  for (int i = 0; i < 10; i++) {
    if (table[i][0] == 0) {
      if (table[i][1] == 0) {
        ando -= curbit;
      }
    } else {
      if (table[i][1] == 0) {
        xoro += curbit;
      } else {
        oro += curbit;
      }
    }
    curbit <<= 1;
  }
  cout << 3 << endl;
  cout << "& " << ando << endl;
  cout << "| " << oro << endl;
  cout << "^ " << xoro << endl;
  return 0;
}
