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;
int N = 10;
bool getBit(long long num, long long idx) {
  return (((num >> idx) & 1LL) == 1);
}
long long clearBit(long long num, long long idx) { return (num & ~(1 << idx)); }
long long setBit(long long num, long long idx) { return (num | (1LL << idx)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, a = (1 << N) - 1, o = 0, xr = 0;
  cin >> n;
  vector<int> ans(N, 2);
  while (n--) {
    char c;
    int x;
    cin >> c >> x;
    for (int i = 0; i < N; ++i) {
      bool parity = getBit(x, i);
      if (c == '&') {
        if (parity == 0) ans[i] = 0;
      } else if (c == '|') {
        if (parity) ans[i] = 1;
      } else {
        if (parity) {
          if (ans[i] == 1 || ans[i] == 0)
            ans[i] = 1 - ans[i];
          else
            ans[i] = 5 - ans[i];
        }
      }
    }
  }
  for (int i = 0; i < ans.size(); ++i) {
    if (ans[i] == 0)
      a = clearBit(a, i);
    else if (ans[i] == 1)
      o = setBit(o, i);
    else if (ans[i] == 3)
      xr = setBit(xr, i);
  }
  cout << 3 << '\n';
  cout << "& " << a << '\n';
  cout << "| " << o << '\n';
  cout << "^ " << xr;
  return 0;
}