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;
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
template <typename T>
T mod_exp(T b, T p, T m) {
  T x = 1;
  while (p) {
    if (p & 1) x = (x * b) % m;
    b = (b * b) % m;
    p = p >> 1;
  }
  return x;
}
template <typename T>
T invFermat(T a, T p) {
  return mod_exp(a, p - 2, p);
}
template <typename T>
T exp(T b, T p) {
  T x = 1;
  while (p) {
    if (p & 1) x = (x * b);
    b = (b * b);
    p = p >> 1;
  }
  return x;
}
const int MAXN = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  int c1 = 1023;
  int c2 = 0;
  int c3 = 0;
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    int val;
    cin >> val;
    if (c == '|') {
      c2 |= val;
      c3 &= ~val;
    } else if (c == '&') {
      c1 &= val;
      c2 &= val;
      c3 &= val;
    } else {
      c3 ^= val;
    }
  }
  cout << 3 << endl;
  cout << "& " << c1 << endl;
  cout << "| " << c2 << endl;
  cout << "^ " << c3 << endl;
  return 0;
}
