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 N = 100005;
int dr8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dc8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dr4[] = {0, 1, -1, 0};
int dc4[] = {1, 0, 0, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  int on = 1023;
  cin >> n;
  int z = 0;
  string s;
  for (int i = 0; i < n; i++) {
    cin >> s;
    int w;
    cin >> w;
    if (s[0] == '|') {
      z = z | w;
      on = on | w;
    } else if (s[0] == '^') {
      z = z ^ w;
      on = on ^ w;
    } else {
      z = z & w;
      on = on & w;
    }
  }
  int a[10];
  int x1 = 0, x2 = 1023, x3 = 0;
  for (int i = 0; i < 10; i++) {
    if (z % 2 == on % 2 && on % 2 == 1) {
      a[i] = 1;
      x1 = x1 ^ (1 << i);
    } else if (z % 2 == on % 2 && on % 2 == 0) {
      a[i] = 0;
      x2 = x2 ^ (1 << i);
    } else if (z % 2 == 1 && on % 2 == 0)
      x3 = x3 ^ (1 << i);
    z = z / 2;
    on = on / 2;
  }
  cout << 3 << endl;
  cout << "| " << x1 << endl;
  cout << "& " << x2 << endl;
  cout << "^ " << x3 << endl;
}