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 main() {
  int i, n, j, num, curr;
  char c;
  std::vector<pair<char, int> > v;
  vector<int> ans;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> c >> num;
    v.push_back({c, num});
  }
  for (i = 0; i < 10; i++) {
    curr = 2;
    for (j = 0; j < n; j++) {
      num = v[j].second;
      c = v[j].first;
      if (c == '|' && (num & (1 << i))) curr = 1;
      if (c == '&' && (num & (1 << i)) == 0) curr = 0;
      if (c == '^' && (num & (1 << i))) {
        if (curr == 2)
          curr = -2;
        else if (curr == -2)
          curr = 2;
        else if (curr == 1)
          curr = 0;
        else if (curr == 0)
          curr = 1;
      }
    }
    ans.push_back(curr);
  }
  int p = 1;
  int xxor = 0, aand = 0, oor = 0;
  for (i = 0; i < 10; i++) {
    if (ans[i] == -2) xxor += p;
    if (ans[i] != 0) aand += p;
    if (ans[i] == 1) oor += p;
    p *= 2;
  }
  cout << "3\n";
  cout << "^ " << xxor << endl;
  cout << "& " << aand << endl;
  cout << "| " << oor << endl;
  return 0;
}
