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 MAX = 1e9 + 10;
const double eps = 1e-4;
const double mod = 1e9 + 7;
bool cal(int a, int i) {
  if (a & (1 << i)) return 1;
  return 0;
}
int main() {
  int n;
  cin >> n;
  int x = 0, y = 1023;
  while (n--) {
    char c[2];
    int t;
    scanf("%s%d", c, &t);
    if (c[0] == '|') x |= t, y |= t;
    if (c[0] == '&') x &= t, y &= t;
    if (c[0] == '^') x ^= t, y ^= t;
  }
  int v1 = 0, v2 = 0, v3 = 0;
  for (int i = 0; i <= 9; i++) {
    if (!cal(x, i) && cal(y, i)) v2 += (1 << i);
    if (cal(x, i) && cal(y, i)) v3 += (1 << i);
    if (cal(x, i) && !cal(y, i)) v2 += (1 << i), v3 += (1 << i);
  }
  printf("3\n");
  printf("| %d\n", v1);
  printf("& %d\n", v2);
  printf("^ %d\n", v3);
  return 0;
}