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 long long int INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int MAX = 1e5 + 5;
int n, d;
char ch;
int ans[10];
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  scanf("%d", &n);
  for (int i = 0; i < 10; i++) ans[i] = 2;
  for (int i = 0; i < n; i++) {
    scanf(" %c %d", &ch, &d);
    bitset<10> temp(d);
    if (ch == '&') {
      for (int i = 0; i < 10; i++) {
        if (temp[i] == 0) ans[i] = 0;
      }
    } else if (ch == '|') {
      for (int i = 0; i < 10; i++) {
        if (temp[i] == 1) ans[i] = 1;
      }
    } else {
      for (int i = 0; i < 10; i++) {
        if (temp[i] == 1) {
          if (ans[i] == 2)
            ans[i] = -1;
          else if (ans[i] == -1)
            ans[i] = 2;
          else
            ans[i] = 1 - ans[i];
        }
      }
    }
  }
  printf("3\n");
  int s = 1023;
  for (int i = 0; i < 10; i++) {
    if (ans[i] == 0) s = s - (1 << i);
  }
  printf("& %d\n", s);
  s = 0;
  for (int i = 0; i < 10; i++) {
    if (ans[i] == 1) s += (1 << i);
  }
  printf("| %d\n", s);
  s = 0;
  for (int i = 0; i < 10; i++) {
    if (ans[i] == -1) s += (1 << i);
  }
  printf("^ %d", s);
  return 0;
}
