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;
int n, k, x;
double l, v1, v2;
string s, op;
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
int exec(int n, string op, int val) {
  if (op == "|") return n | val;
  if (op == "^") return n ^ val;
  if (op == "&") return n & val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int x0 = 0, x1 = 1023;
  for (int i = 1; i <= n; i++) {
    cin >> op >> x;
    x0 = exec(x0, op, x);
    x1 = exec(x1, op, x);
  }
  int xxor = 0;
  int aand = 1023;
  int oor = 0;
  for (int i = 0; i < 10; i++) {
    if (x1 & (1 << i) && !(x0 & (1 << i))) continue;
    if (!(x1 & (1 << i)) && (x0 & (1 << i))) {
      xxor += (1 << i);
    }
    if (!(x1 & (1 << i)) && !(x0 & (1 << i))) {
      aand -= (1 << i);
    }
    if ((x1 & (1 << i)) && (x0 & (1 << i))) {
      oor += (1 << i);
    }
  }
  cout << "3\n";
  cout << "| " << oor << '\n';
  cout << "& " << aand << '\n';
  cout << "^ " << xxor << '\n';
}
