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 int maxn = 1200;
bool fix[maxn];
int val[maxn], xo[maxn];
int main() {
  int n, x;
  char s[10];
  while (scanf("%d", &n) == 1) {
    memset(fix, 0, sizeof(fix));
    memset(xo, 0, sizeof(xo));
    memset(val, 0, sizeof(val));
    for (int i = 0; i < n; i++) {
      scanf("%s%d", s, &x);
      for (int j = 0; j < 10; j++) {
        int t = (x & (1 << j)) ? 1 : 0;
        if (fix[j]) {
          if (s[0] == '&')
            val[j] &= t;
          else if (s[0] == '|')
            val[j] |= t;
          else
            val[j] ^= t;
        } else {
          if (s[0] == '&') {
            if (t == 0) {
              fix[j] = 1;
              val[j] = 0;
            }
          } else if (s[0] == '|') {
            if (t == 1) {
              fix[j] = 1;
              val[j] = 1;
            }
          } else {
            xo[j] ^= t;
          }
        }
      }
    }
    puts("3");
    int num = 1023;
    for (int j = 0; j < 10; j++)
      if (fix[j] && val[j] == 0) num -= (1 << j);
    printf("& %d\n", num);
    num = 0;
    for (int j = 0; j < 10; j++)
      if (fix[j] && val[j] == 1) num += (1 << j);
    printf("| %d\n", num);
    num = 0;
    for (int j = 0; j < 10; j++)
      if (!fix[j] && xo[j]) num += (1 << j);
    printf("^ %d\n", num);
  }
  return 0;
}
