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 t, n;
int pre[20];
int sat[20], x;
char s[20];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    scanf("%d", &x);
    if (s[0] == '|') {
      for (int i = 0; x; x >>= 1, i++)
        if (x & 1) {
          sat[i] = 1;
        }
    } else if (s[0] == '&') {
      memset(pre, 0, sizeof pre);
      for (int i = 0; x; x >>= 1, i++)
        if (x & 1) pre[i] = 1;
      for (int i = 0; i < 11; i++)
        if (pre[i] == 0) sat[i] = 2;
    } else {
      for (int i = 0; x; x >>= 1, i++)
        if (x & 1) {
          if (sat[i] == 0) {
            sat[i] = 3;
          } else {
            sat[i] = 3 - sat[i];
          }
        }
    }
  }
  int k1 = 0, k2 = 0, k3 = 0;
  for (int i = 0; i < 10; i++)
    if (sat[i] != 2) k1 += (1 << i);
  for (int i = 0; i < 10; i++)
    if (sat[i] == 1) k2 += (1 << i);
  for (int i = 0; i < 10; i++)
    if (sat[i] == 3) k3 += (1 << i);
  int cnt = 0;
  if (k1 != 1023) cnt++;
  if (k2 != 0) cnt++;
  if (k3 != 0) cnt++;
  printf("%d", cnt);
  if (k1 != 1023) printf("\n& %d", k1);
  if (k2 != 0) printf("\n| %d", k2);
  if (k3 != 0) printf("\n^ %d", k3);
}
