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 SIZE = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const double PI = atan(1.0) * 4.0;
const long long mod = 1e9 + 7;
int n, ans;
char opt[2];
int d[12] = {0};
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  for (int i = 1, x; i <= n; i++) {
    cin >> opt >> x;
    if (opt[0] == '|') {
      for (int j = 0; j < 10; j++)
        if ((x >> j) & 1) d[j] = 1;
      continue;
    }
    if (opt[0] == '^') {
      for (int j = 0; j < 10; j++)
        if ((x >> j) & 1) {
          if (d[j] == 1)
            d[j] = -1;
          else if (d[j] == -1)
            d[j] = 1;
          else if (d[j] == 0)
            d[j] = 2;
          else
            d[j] = 0;
        }
      continue;
    }
    for (int j = 0; j < 10; j++)
      if (((x >> j) & 1) == 0) d[j] = -1;
  }
  int v = 1, c = 0;
  for (int i = 0; i < 10; i++) {
    if (d[i] != 0) v = 0;
    if (d[i] == 1 || d[i] == -1) c |= (1 << i);
  }
  if (v) {
    cout << 0;
    return 0;
  }
  cout << 2 << "\n";
  cout << "| " << c << "\n";
  c = 0;
  for (int i = 0; i < 10; i++)
    if (d[i] == -1 || d[i] == 2) c |= (1 << i);
  cout << "^ " << c;
  return 0;
}
