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 double eps = 0.000000000000001;
string convertstring(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
int a[11][2];
bool bit(int a, int in) {
  if (a & (1 << in)) return 1;
  return 0;
}
int main() {
  int n, k, i, j;
  scanf("%d", &n);
  ;
  for (i = 0; i < 10; ++i)
    for (j = 0; j < 2; ++j) a[i][j] = j;
  getchar();
  for (i = 0; i < n; ++i) {
    char c, s[3];
    int x;
    scanf("%s%d", s, &x);
    c = s[0];
    for (j = 0; j < 10; ++j)
      for (k = 0; k < 2; ++k) {
        if (c == '|')
          a[j][k] |= (bit(x, j));
        else if (c == '&')
          a[j][k] &= (bit(x, j));
        else if (c == '^')
          a[j][k] ^= (bit(x, j));
      }
  }
  int ann = 0, orr = 0, xrr = 0;
  for (i = 0; i < 10; ++i) {
    if (!a[i][0] && !a[i][1]) {
    } else if (a[i][0] && a[i][1]) {
      orr |= (1 << i);
      ann |= (1 << i);
    } else if (a[i][0] && !a[i][1]) {
      xrr |= (1 << i);
      ann |= (1 << i);
    } else
      ann |= (1 << i);
  }
  printf("3\n& %d\n| %d\n^ %d\n", ann, orr, xrr);
  return 0;
}