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;
struct oper {
  char a;
  short int b;
};
int n, newn;
int a = 1023, o = 0, x = 0;
int r1 = 1023, r2 = 0;
vector<oper> ops, newops;
void Input() {
  cin >> n;
  ops.resize(n);
  for (int i = 0; i < n; i++) cin >> ops[i].a >> ops[i].b;
}
void Body() {
  for (int i = 0; i < n; i++) {
    if (ops[i].a == '|') {
      r1 = r1 | ops[i].b;
      r2 = r2 | ops[i].b;
    }
    if (ops[i].a == '&') {
      r1 = r1 & ops[i].b;
      r2 = r2 & ops[i].b;
    }
    if (ops[i].a == '^') {
      r1 = r1 ^ ops[i].b;
      r2 = r2 ^ ops[i].b;
    }
  }
  int q1, q2, tmp;
  for (int i = 0; i < 10; i++) {
    q1 = r1 % 2;
    r1 = r1 / 2;
    q2 = r2 % 2;
    r2 = r2 / 2;
    tmp = pow((float)2, i);
    if (q1 == q2) {
      if (q1 == 0)
        a = a & ~tmp;
      else
        o = o | tmp;
    } else if (q1 == 0)
      x = x | tmp;
  }
}
void Output() {
  if (a != 1023) newn++;
  if (o != 0) newn++;
  if (x != 0) newn++;
  cout << newn << endl;
  if (a != 1023) cout << "& " << a << endl;
  if (o != 0) cout << "| " << o << endl;
  if (x != 0) cout << "^ " << x << endl;
}
int main() {
  Input();
  Body();
  Output();
  return 0;
}