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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int N = 500005;
int n, val[N], type_[N], bitval[12];
char op[5];
int get_bit(int v, int p) { return (v >> p) & 1; }
int main() {
  scanf("%d", &n);
  int xorval = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%s%d", op, &val[i]);
    if (op[0] == '|') {
      type_[i] = 0;
    } else if (op[0] == '&') {
      type_[i] = 1;
    } else {
      type_[i] = 2;
      xorval ^= val[i];
    }
  }
  for (int b = 0; b < 10; b++) {
    bitval[b] = -1;
    for (int i = 1; i <= n; i++) {
      if (type_[i] == 0 && get_bit(val[i], b) == 1)
        bitval[b] = 1;
      else if (type_[i] == 1 && get_bit(val[i], b) == 0)
        bitval[b] = 0;
      else if (type_[i] == 2 && get_bit(val[i], b) == 1 && bitval[b] != -1)
        bitval[b] ^= 1;
    }
  }
  printf("2\n");
  int orval = 0;
  for (int i = 0; i < 10; i++)
    if (bitval[i] == 1) {
      orval |= (1 << i);
      xorval &= 1023 ^ (1 << i);
    } else if (bitval[i] == 0) {
      orval |= (1 << i);
      xorval |= (1 << i);
    }
  printf("| %d\n", orval);
  printf("^ %d\n", xorval);
  return 0;
}