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 = 5e5 + 100;
int mark[maxn], num[maxn];
int main(void) {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      char op[5];
      int type;
      scanf("%s%d", op, &num[i]);
      if (op[0] == '|')
        type = 1;
      else if (op[0] == '&')
        type = 2;
      else if (op[0] == '^')
        type = 3;
      mark[i] = type;
    }
    int res1 = 0, res2 = 1023, res3 = 0;
    for (int i = 0; i < 10; i++) {
      int type = 0, tou = 1;
      for (int j = 1; j <= n; j++) {
        if (num[j] & (1 << i)) {
          if (mark[j] == 1) {
            type = 1;
            tou = j + 1;
          }
        } else if (mark[j] == 2) {
          type = 2;
          tou = j + 1;
        }
      }
      if (type == 1)
        res1 += (1 << i);
      else if (type == 2)
        res2 -= (1 << i);
      int flag = 0;
      for (int j = tou; j <= n; j++) {
        if (num[j] & (1 << i)) {
          if (mark[j] == 3) flag ^= 1;
        }
      }
      if (flag) res3 += (1 << i);
    }
    puts("3");
    printf("| %d\n& %d\n^ %d\n", res1, res2, res3);
  }
  return 0;
}
