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 = 5e5;
char op[maxn];
int x[maxn];
int ansx[maxn], anso[maxn], ansa[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> op[i] >> x[i];
  for (int i = 0; i < 10; i++) {
    int t1 = 0;
    for (int j = 0; j < n; j++) {
      if (op[j] == '|' && ((1 << i) & x[j])) t1 = t1 | 1;
      if (op[j] == '^' && ((1 << i) & x[j])) t1 = t1 ^ 1;
      if (op[j] == '&' && !((1 << i) & x[j])) t1 = 0;
    }
    int t2 = 1;
    for (int j = 0; j < n; j++) {
      if (op[j] == '|' && ((1 << i) & x[j])) t2 = t2 | 1;
      if (op[j] == '^' && ((1 << i) & x[j])) t2 = t2 ^ 1;
      if (op[j] == '&' && !((1 << i) & x[j])) t2 = 0;
    }
    if (t1 == 0 && t2 == 0) ansx[i] = 0, anso[i] = 0, ansa[i] = 0;
    if (t1 == 0 && t2 == 1) ansx[i] = 0, anso[i] = 0, ansa[i] = 1;
    if (t1 == 1 && t2 == 0) ansx[i] = 1, anso[i] = 0, ansa[i] = 1;
    if (t1 == 1 && t2 == 1) ansx[i] = 0, anso[i] = 1, ansa[i] = 1;
  }
  cout << "3\n";
  int Ax = 0;
  for (int i = 0; i < 10; i++)
    if (ansx[i]) Ax += (1 << i);
  int Ao = 0;
  for (int i = 0; i < 10; i++)
    if (anso[i]) Ao += (1 << i);
  int Aa = 0;
  for (int i = 0; i < 10; i++)
    if (ansa[i]) Aa += (1 << i);
  cout << "^ " << Ax << endl;
  cout << "& " << Aa << endl;
  cout << "| " << Ao << endl;
  return 0;
}