443 C. Short Program

http://codeforces.com/contest/879/problem/C

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

Copy

3
| 3
^ 2
| 1

output

2
| 3
^ 2

input

Copy

3
& 1
& 3
& 5

output

1
& 1

input

Copy

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.

题意：给出一段程序，只有三种操作，^ & |，然后让你输出一段不超过5行的程序使得运算结果和给出程序的运算结果相同

这题注意几点：

1.母数用000000不够，还需要与1111111结合，将其经过处理后数的每一位发生变化 0->? 1->?,

所以对于每个处理的数位我们有：

（0->0,1->0）none

（0->1,1->1）or 1

  (0->1,1->0)  xor 1

  (0->0,1->1)  and 1

2.位运算处理数后保持原样的操作有这三个  & 1    | 0   ^ 0 

3.复习下位运算的优先级 AND>XOR>OR

#include <bits/stdc++.h>
using namespace std;
#define maxn 100000
typedef long long ll;
#define inf 2147483647
#define ri register int

int n, x;
char ch;
int f0 = 0, f1 = 1023;
int AND = 0, XOR = 0, OR = 0;

int main() {
  ios::sync_with_stdio(false);
  // freopen("test.txt", "r", stdin);
  //  freopen("outout.txt","w",stdout);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ch >> x;
    if (ch == '|') {
      f0 |= x;
      f1 |= x;
    }
    if (ch == '&') {
      f0 &= x;
      f1 &= x;
    }
    if (ch == '^') {
      f0 ^= x;
      f1 ^= x;
    }
  }
  int base = 1;
  int w0, w1;

  for (int i = 1; i <= 10; i++) {
    w0 = f0 & 1;
    w1 = f1 & 1;
    if (w0 == 0) {
      if (w1 == 1)
        AND += base;
    } else {
      if (w1 == 0) {
        AND += base;
        XOR += base;
      } else {
        AND += base;
        OR += base;
      }
    }
    base <<= 1;
    f0 >>= 1, f1 >>= 1;
  }
  cout << 3 << endl
       << "& " << AND << endl
       << "^ " << XOR << endl
       << "| " << OR << endl;

  return 0;
}