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 mod = 1e9 + 7;
const int N = 5e5 + 10;
int n;
char sign[N];
int val[N];
int Or = 0, And = 0, Xor = 0;
int giveres(int v) {
  for (int i = 0; i < n; i++) {
    if (sign[i] == '|')
      v = v | val[i];
    else if (sign[i] == '^')
      v = v ^ val[i];
    else
      v = v & val[i];
  }
  return v;
}
int giveres2(int v) {
  v = v | Or;
  v = v & And;
  v = v ^ Xor;
  return v;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> sign[i] >> val[i];
  vector<int> oo(20), aa(20), xx(20);
  int res2 = giveres(0);
  for (int i = 0; i < 10; i++) {
    int sm = (1 << i);
    int res1 = giveres(sm);
    int bt1 = (res1 >> i) & 1;
    int bt2 = (res2 >> i) & 1;
    if (bt1 == 0 && bt2 == 0) {
      aa[i] = 1;
    } else if (bt1 == 0 && bt2 == 1) {
      xx[i] = 1;
      Xor = Xor | (1 << i);
    } else if (bt1 == 1 && bt2 == 0) {
      aa[i] = 1;
      And = And | (1 << i);
    } else {
      oo[i] = 1;
      Or = Or | (1 << i);
    }
  }
  for (int i = 0; i < 10; i++) {
    if (!aa[i]) {
      And = And | (1 << i);
    }
  }
  cout << 3 << "\n";
  cout << "& " << And << "\n"
       << "^ " << Xor << "\n"
       << "| " << Or << "\n";
  return 0;
}