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 = 10;
const int inf = 0x3f3f3f3f;
int num[N];
int xx[N];
void f_or(int x) {
  int i = 0;
  while (x) {
    if (x % 2) {
      num[i] = 1;
    }
    i++;
    x /= 2;
  }
}
void f_and(int x) {
  int i = 0;
  int safe[10];
  memset(safe, 0, sizeof safe);
  while (x) {
    if (x % 2) {
      safe[i] = 1;
    }
    i++;
    x /= 2;
  }
  for (int i = 0; i < N; i++) {
    if (!safe[i]) {
      num[i] = 0;
    }
  }
  cout << endl;
}
void f_xor(int x) {
  int i = 0;
  while (x) {
    if (x % 2) {
      xx[i] = !xx[i];
      if (num[i] == 1)
        num[i] = 0;
      else if (num[i] == 0)
        num[i] = 1;
    }
    i++;
    x /= 2;
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  memset(num, -1, sizeof num);
  for (int i = 0; i < n; i++) {
    char c;
    int y;
    cin >> c >> y;
    cin.ignore();
    if (c == '|') {
      f_or(y);
    } else if (c == '^') {
      f_xor(y);
    } else {
      f_and(y);
    }
  }
  int a = 1023, o = 0, x = 0;
  for (int i = 0; i < N; i++) {
    if (num[i] == 1) o += (1 << i);
  }
  for (int i = 0; i < N; i++) {
    if (num[i] == 0) a -= (1 << i);
  }
  for (int i = 0; i < N; i++) {
    if (num[i] == -1 && xx[i]) x += (1 << i);
  }
  cout << 3 << endl;
  cout << "& " << a << endl;
  cout << "| " << o << endl;
  cout << "^ " << x << endl;
  return 0;
}