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;
char op[1000000];
int var[1000000];
int pos[729];
int main() {
  int x = 0;
  int y = 1023;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf(" %c %d", &op[i], &var[i]);
    if (op[i] == '&') {
      x &= var[i];
      y &= var[i];
    } else if (op[i] == '|') {
      x |= var[i];
      y |= var[i];
    } else {
      x ^= var[i];
      y ^= var[i];
    }
  }
  memset(pos, -1, sizeof(pos));
  for (int i = 0; i < 729; i++) {
    bool foundall = true;
    vector<int> ans;
    for (int ori = 0; ori < 10; ori++) {
      bool found = false;
      for (int j = 0; j < 32; j++) {
        int ops = i;
        int one = 1;
        int zero = 0;
        for (int c = 0; c < 5; c++) {
          int cbit = ((j >> c) & 1);
          if (ops % 3 == 0) {
            one &= cbit;
            zero &= cbit;
          } else if (ops % 3 == 1) {
            one |= cbit;
            zero |= cbit;
          } else {
            one ^= cbit;
            zero ^= cbit;
          }
          ops /= 3;
        }
        if (((x >> ori) & 1) == zero && ((y >> ori) & 1) == one) {
          ans.push_back(j);
          found = true;
          break;
        }
      }
      if (found == false) {
        foundall = false;
        break;
      }
    }
    if (foundall) {
      cout << 5 << endl;
      int ops = i;
      string opc = "&|^";
      for (int j = 0; j < 5; j++) {
        int cnum = 0;
        for (int c = 0; c < 10; c++) {
          cnum |= (((ans[c] >> j) & 1) << c);
        }
        printf("%c %d\n", opc[ops % 3], cnum);
        ops /= 3;
      }
      return 0;
    }
  }
  return 0;
}