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;
long long int arr[1000003], pre[1000003] = {0}, dp[1000003], dp1[1000003],
                            dip[1000003] = {0};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, i;
  cin >> n;
  long long int z = 0, o = 1023;
  for (i = 0; i < n; i++) {
    char c;
    int m;
    cin >> c >> m;
    if (c == '^') {
      z = z ^ m;
      o = o ^ m;
    } else if (c == '&')
      z &= m, o &= m;
    else
      z |= m, o |= m;
  }
  int a1 = 1023, a2 = 0, a3 = 1023, a4 = 0;
  int p = 1;
  for (i = 0; i < 10; i++) {
    if (z % 2 == 0 and o % 2 == 0) {
      a1 -= p;
    } else if (z % 2 == 1 and o % 2 == 1) {
      a2 += p;
    } else if (z % 2 == 0 and o % 2 == 1) {
      a3 -= p;
    } else {
      a4 += p;
    }
    z /= 2;
    o /= 2;
    p *= 2;
  }
  cout << 3 << endl;
  cout << "& " << a1 << endl;
  cout << "| " << a2 << endl;
  cout << "^ " << a4 << endl;
  return 0;
}