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;
using INT = long long;
const int NN = 5e5 + 10;
char s[NN];
int a[NN];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf(" %c %d", s + i, a + i);
  }
  int sor = 0, sand = (1 << 10) - 1, sxor = 0;
  for (int i = 0; i < 10; i++) {
    int st1 = (1 << i);
    int st0 = 0;
    for (int j = 1; j <= n; j++) {
      int x = 0;
      if ((a[j] >> i) & 1) x = (1 << i);
      if (s[j] == '|') st1 |= x, st0 |= x;
      if (s[j] == '&') st1 &= x, st0 &= x;
      if (s[j] == '^') st1 ^= x, st0 ^= x;
    }
    if (st1 and st0) {
      sor ^= (1 << i);
    }
    if (!st1 and st0) {
      sxor ^= (1 << i);
    }
    if (!st1 and !st0) {
      sand ^= (1 << i);
    }
  }
  cout << 3 << endl;
  cout << "|" << ' ' << sor << endl;
  cout << "&" << ' ' << sand << endl;
  cout << "^" << ' ' << sxor << endl;
}
