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;
using ll = long long;
const auto R_ = [] { return ios::sync_with_stdio(false), cin.tie(nullptr); }();
constexpr int LG = 10;
int change[LG];
signed main() {
  int n;
  scanf("%d", &n);
  fill(change, change + LG, 10);
  for (int i = 0; i < n; ++i) {
    char c;
    int x;
    scanf(" %c %d", &c, &x);
    if (c == '|')
      for (int i = 0; i < LG; ++i)
        if ((x >> i) & 1) change[i] = 1;
    if (c == '&')
      for (int i = 0; i < LG; ++i)
        if (!((x >> i) & 1)) change[i] = 0;
    if (c == '^')
      for (int i = 0; i < LG; ++i)
        if ((x >> i) & 1) change[i] ^= 1;
  }
  int and_ = 0, or_ = 0, xor_ = 0;
  for (int i = 0; i < LG; ++i) {
    if (change[i] == 0) and_ |= 1 << i;
    if (change[i] == 1) or_ |= 1 << i;
    if (change[i] == 11) xor_ |= 1 << i;
  }
  printf("3\n| %d\n& %d\n^ %d", or_, 1023 ^ and_, xor_);
  return 0;
}