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 + 233;
int v1 = 0, v2 = 0, v3 = 0;
int n;
int op[maxn], v[maxn];
int cal(int d, int x) {
  for (int i = 1; i <= n; i++) {
    int t = (v[i] >> d) & 1;
    if (op[i] == 0)
      x &= t;
    else if (op[i] == 1)
      x |= t;
    else
      x ^= t;
  }
  return x;
}
int main() {
  cin >> n;
  getchar();
  for (int i = 1; i <= n; i++) {
    char c = getchar();
    if (c == '&')
      op[i] = 0;
    else if (c == '|')
      op[i] = 1;
    else
      op[i] = 2;
    scanf("%d", &v[i]);
    getchar();
  }
  for (int i = 0; i < 10; i++) {
    if (cal(i, 0) == 0) {
      if (cal(i, 1) == 0) {
      } else {
        v2 |= (1 << i);
      }
    } else {
      if (cal(i, 1) == 0) {
        v2 |= (1 << i);
        v3 |= (1 << i);
      } else {
        v1 |= (1 << i);
        v2 |= (1 << i);
      }
    }
  }
  cout << 3 << endl;
  printf("| %d\n& %d\n^ %d\n", v1, v2, v3);
  return 0;
}
