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 N = 5e5 + 5;
int n;
pair<int, int> p[N];
int a, o, x;
inline int trans(int cur) {
  for (int i = (0); i < (n); ++i) {
    if (p[i].first == 0)
      cur ^= p[i].second;
    else if (p[i].first == 1)
      cur |= p[i].second;
    else
      cur &= p[i].second;
  }
  return cur;
}
int main() {
  scanf("%d", &n);
  char c[4];
  for (int i = (0); i < (n); ++i) {
    scanf("%s", c);
    if (c[0] == '^')
      p[i].first = 0;
    else if (c[0] == '|')
      p[i].first = 1;
    else
      p[i].first = 2;
    scanf("%d", &p[i].second);
  }
  for (int i = (0); i < (10); ++i) {
    int d1 = (trans(1 << i) >> i) & 1;
    int d0 = (trans(0) >> i) & 1;
    if (d1 == 1) {
      if (d0 == 1)
        o ^= 1 << i, a ^= 1 << i;
      else if (d0 == 0)
        a ^= 1 << i;
    } else {
      if (d0 == 1)
        a ^= 1 << i, x ^= 1 << i;
      else if (d0 == 0)
        o ^= 1 << i, x ^= 1 << i;
    }
  }
  printf("%d\n& %d\n| %d\n^ %d\n", 3, a, o, x);
  return 0;
}
