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 N = 500005;
int n, a[N];
char c[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> c[i] >> a[i];
  }
  int x, y, z;
  z = 1023, x = y = 0;
  for (int i = 0; i < 10; ++i) {
    int pos = -1;
    bool t = 0;
    for (int j = 0; j < n; ++j) {
      if (c[j] == '|' && a[j] >> i & 1) pos = j, t = 1;
      if (c[j] == '&' && !(a[j] >> i & 1)) pos = j, t = 0;
    }
    for (int j = pos + 1; j < n; ++j) {
      if (c[j] == '^') t ^= (a[j] >> i & 1);
    }
    if (pos == -1)
      x |= t * 1 << i;
    else {
      if (t)
        y |= 1 << i;
      else
        z ^= 1 << i;
    }
  }
  cout << "3\n";
  cout << "^ " << x << '\n';
  cout << "| " << y << '\n';
  cout << "& " << z << '\n';
}
