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;
using ii = tuple<long long, long long>;
const long long N = 1e6 + 7, inf = 1e9 + 42;
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return (res * res * a) % 1000000007;
  else
    return (res * res) % 1000000007;
}
bool compare(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first)
    return (a.second > b.second);
  else
    return (a.first <= b.first);
}
char op[(long long)1e6];
long long A[(long long)1e6];
signed main() {
  long long n;
  cin >> n;
  ;
  long long P = 0, S = 0;
  for (long long i = 1; i < n + 1; i++) {
    cin >> op[i] >> A[i];
    if (op[i] == '|') {
      P |= A[i];
      S |= A[i];
    }
    if (op[i] == '&') {
      P &= A[i];
      S |= (1023 - A[i]);
    }
    if (op[i] == '^') P ^= A[i];
  }
  cout << "2" << endl;
  cout << "| " << S << endl;
  cout << "^ " << (P ^ S) << endl;
}
