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;
const int maxn = 1e5 + 10;
const int maxm = 1e3 + 5;
const int mod = 1e9 + 7;
inline void solve() {
  int n, mask = 0, b1 = 0, b2 = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    char op[2];
    int x;
    scanf("%s%d", op, &x);
    if (op[0] == '|') {
      for (int j = 0; j < 10; j++)
        if (x & (1 << j)) {
          mask |= (1 << j);
          b1 |= (1 << j);
        }
    } else if (op[0] == '&') {
      for (int j = 0; j < 10; j++)
        if ((x & (1 << j)) == 0) {
          mask |= 1 << j;
          b1 &= (1 << 10) - 1 - (1 << j);
        }
    } else if (op[0] == '^') {
      for (int j = 0; j < 10; j++)
        if (x & (1 << j)) {
          if (mask & (1 << j)) b1 ^= (1 << j);
          b2 ^= (1 << j);
        }
    }
  }
  puts("3");
  int ans1 = 1023, ans2 = 0, ans3 = 0;
  for (int i = 0; i < 10; i++)
    if (mask & (1 << i)) {
      if (b1 & (1 << i))
        ans2 |= (1 << i);
      else
        ans1 ^= (1 << i);
    } else if (b2 & (1 << i))
      ans3 |= (1 << i);
  printf("& %d\n| %d\n^ %d\n", ans1, ans2, ans3);
}
int main() {
  solve();
  return 0;
}