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 long long INF = 1e9 + 10;
const long double PI = acos(-1);
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
const int M = 1003;
const int N = 1e6 + 10;
int a[N];
int main() {
  int n;
  cin >> n;
  int x1 = 0, x2 = 1023;
  for (int i = 0; i < (int)n; i++) {
    char ch;
    cin >> ch;
    int x;
    cin >> x;
    if (ch == '|') {
      x1 |= x;
      x2 |= x;
    } else if (ch == '^') {
      x1 ^= x;
      x2 ^= x;
    } else {
      x1 &= x;
      x2 &= x;
    }
  }
  int and1 = 1023, or1 = 0, xor1 = 0;
  for (int i = 0; i < (int)10; i++) {
    int m = (1 << i);
    if ((m & x1) && (m & x2)) {
      or1 |= m;
    } else if (!(m & x1) && !(m & x2)) {
      and1 ^= m;
    } else if (m & x1) {
      xor1 |= m;
    }
  }
  cout << 3 << endl;
  cout << "| " << or1 << endl;
  cout << "^ " << xor1 << endl;
  cout << "& " << and1 << endl;
}