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 N = 5e5 + 9;
typedef struct {
  char c;
  long long x;
  void read() { cin >> c >> x; }
  void run(long long& prev, long long i) {
    long long mask = 1LL << i;
    long long bit = min(1LL, mask & x);
    if (c == '|')
      prev |= bit;
    else if (c == '&')
      prev &= bit;
    else
      prev ^= bit;
  }
} operation;
operation op[N];
long long value(vector<long long>& a) {
  long long ret = 0;
  for (long long i = 0; i < a.size(); i++) {
    ret += (long long)(pow(2, i) + 0.5) * a[i];
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) op[i].read();
  vector<long long> aand(10, 1), oor(10), xxor(10);
  for (int i = 0; i < 10; i++) {
    long long cng = 0;
    vector<long long> val;
    for (long long j = 0; j < 2; j++) {
      long long prev = j;
      for (int k = 0; k < n; k++) op[k].run(prev, i);
      if (prev != j) cng++;
      val.push_back(prev);
    }
    if (cng == 2)
      xxor[i] = 1;
    else if (cng == 1) {
      if (val[0] == 0)
        aand[i] = 0;
      else
        oor[i] = 1;
    }
  }
  cout << 3 << "\n";
  cout << "| " << value(oor) << "\n";
  cout << "& " << value(aand) << "\n";
  cout << "^ " << value(xxor) << "\n";
}