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;
const int INF = 0x3f3f3f3f;
const int maxn = 5e5 + 5;
char s[maxn][2];
int d[maxn];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s %d", s[i], &d[i]);
  }
  int number0 = 0, number1 = 1023;
  for (int i = 0; i < n; i++) {
    if (s[i][0] == '|') {
      number0 = number0 | d[i];
      number1 = number1 | d[i];
    } else if (s[i][0] == '^') {
      number0 = number0 ^ d[i];
      number1 = number1 ^ d[i];
    } else {
      number0 = number0 & d[i];
      number1 = number1 & d[i];
    }
  }
  int base = 1, XOR = 0, OR = 0, AND = 0;
  for (int i = 0; i < 10; i++) {
    int low0 = number0 & 1;
    int low1 = number1 & 1;
    if (low0 && !low1) {
      XOR = XOR | base;
    } else if (low0 && low1) {
      OR = OR | base;
    } else if (!low0 && !low1) {
      OR = OR | base;
      XOR = XOR | base;
    }
    base <<= 1;
    number0 >>= 1;
    number1 >>= 1;
  }
  printf("2\n| %d\n^ %d\n", OR, XOR);
  return 0;
}
