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;
const int bits = 10;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<string, int> > commands;
  for (int i = 0; i < n; ++i) {
    string tp;
    int x;
    cin >> tp >> x;
    commands.emplace_back(tp, x);
  }
  int ansxor = 0;
  int ansand = (1 << bits) - 1;
  int ansor = 0;
  for (int bit = 0; bit < bits; ++bit) {
    vector<int> f = {0, 1};
    for (int index = 0; index < 2; ++index) {
      for (int i = 0; i < n; ++i) {
        string tp;
        int x;
        tie(tp, x) = commands[i];
        x = (x >> bit) & 1;
        if (tp == "&") f[index] &= x;
        if (tp == "|") f[index] |= x;
        if (tp == "^") f[index] ^= x;
      }
    }
    if (f == vector<int>{0, 0}) {
      ansand ^= 1 << bit;
    } else if (f == vector<int>{1, 0}) {
      ansxor ^= 1 << bit;
    } else if (f == vector<int>{1, 1}) {
      ansor ^= 1 << bit;
    }
  }
  cout << 3 << endl;
  cout << "^ " << ansxor << endl;
  cout << "& " << ansand << endl;
  cout << "| " << ansor << endl;
}
