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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  while (cin >> n) {
    string s;
    int k;
    int x = 1023, y = 0;
    for (int i = 0; i < n; i++) {
      cin >> s >> k;
      if (s == "^") {
        x ^= k;
        y ^= k;
      } else if (s == "|") {
        x |= k;
        y |= k;
      } else {
        x &= k;
        y &= k;
      }
    }
    int xo = 0, o = 0;
    for (int i = 1; i < 1024; i <<= 1) {
      if ((x & i) && (y & i)) {
        o += i;
      } else if (!(x & i) && !(y & i)) {
        o += i, xo += i;
      } else if (!(x & i) && (y & i)) {
        xo += i;
      }
    }
    cout << 2 << endl;
    cout << "|" << ' ' << o << endl;
    cout << "^" << ' ' << xo << endl;
  }
}