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;
const int N = 500005;
pair<char, int> arr[N];
int bits[10][2];
int main() {
  int n;
  scanf("%d%*c", &n);
  for (int i = 0; i < n; i++) {
    if (i == n - 1)
      scanf("%c %d", &arr[i].first, &arr[i].second);
    else
      scanf("%c %d%*c", &arr[i].first, &arr[i].second);
  }
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 2; j++) {
      int curr = j;
      for (int k = 0; k < n; k++) {
        int currBit = (arr[k].second & (1 << i)) ? 1 : 0;
        if (arr[k].first == '^') {
          curr = curr ^ currBit;
        } else if (arr[k].first == '|') {
          curr = curr | currBit;
        } else {
          curr = curr & currBit;
        }
      }
      bits[i][j] = curr;
    }
  }
  int a[10], b[10], c[10];
  for (int i = 0; i < 10; i++) {
    if (!bits[i][0] && !bits[i][1]) {
      a[i] = 0;
      b[i] = 0;
      c[i] = 0;
    } else if (!bits[i][0] && bits[i][1]) {
      a[i] = 1;
      b[i] = 0;
      c[i] = 0;
    } else if (bits[i][0] && !bits[i][1]) {
      a[i] = 1;
      b[i] = 1;
      c[i] = 0;
    } else {
      a[i] = 1;
      b[i] = 0;
      c[i] = 1;
    }
  }
  int aa = 0, bb = 0, cc = 0;
  for (int i = 0; i < 10; i++)
    aa |= a[i] * (1 << i), bb |= b[i] * (1 << i), cc |= c[i] * (1 << i);
  printf("3\n");
  printf("& %d\n", aa);
  printf("^ %d\n", bb);
  printf("| %d", cc);
  return 0;
}