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 change(int s, char op, int val) {
  switch (op) {
    case '&':
      return val == 0 ? 0 : s;
      break;
    case '|':
      return val == 1 ? 1 : s;
      break;
    case '^':
      return s ^ val;
      break;
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> r(10, 2);
  for (int i = 0; i < n; ++i) {
    char c;
    int v;
    cin >> c >> v;
    for (int j = 0; j < 10; ++j) {
      r[j] = change(r[j], c, v & 1);
      v >>= 1;
    }
  }
  cout << 2 << endl;
  int andOp = 0;
  for (int j = 0; j < 10; ++j) {
    if (r[j] < 2) {
      andOp |= 1 << j;
    }
  }
  andOp = (~andOp) & ((1 << 10) - 1);
  cout << "& " << andOp << endl;
  int xorOp = 0;
  for (int j = 0; j < 10; ++j) {
    if (r[j] & 1) {
      xorOp |= 1 << j;
    }
  }
  cout << "^ " << xorOp << endl;
}
int main() {
  srand(132456);
  solve();
  return 0;
}
