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 maxn = 5e5 + 17, z = 10;
bool ch[z];
char t[maxn];
int x[maxn], n;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> t[i] >> x[i];
  for (int k = 0; k < z; k++) {
    int y = 1 << k;
    for (int i = 0; i < n; i++)
      if (t[i] == '|')
        y |= x[i];
      else if (t[i] == '^')
        y ^= x[i];
      else
        y &= x[i];
    ch[k] = y >> k & 1;
  }
  int base = 0;
  for (int i = 0; i < n; i++)
    if (t[i] == '|')
      base |= x[i];
    else if (t[i] == '^')
      base ^= x[i];
    else
      base &= x[i];
  int And = (1 << 10) - 1, Or = 0, Xor = 0;
  for (int i = 0; i < z; i++)
    if (ch[i] == 1 && (base >> i & 1))
      Or |= 1 << i;
    else if (ch[i] == 0 && (base >> i & 1))
      Xor |= 1 << i;
    else if (ch[i] == 0 && !(base >> i & 1))
      And ^= 1 << i;
  cout << 3 << '\n';
  cout << "& " << And << '\n';
  cout << "| " << Or << '\n';
  cout << "^ " << Xor << '\n';
}
