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.
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 10;
int d[10][2];
int a[MAXN], b[MAXN];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    char s[10];
    scanf("%s%d", s, &b[i]);
    if (s[0] == '|')
      a[i] = 1;
    else if (s[0] == '&')
      a[i] = 2;
    else
      a[i] = 3;
  }
  int a1 = 0, a2 = 0, a3 = 0;
  for (int i = 0; i < 10; i++) {
    int p = 0, q = 1;
    for (int j = 0; j < n; j++) {
      if (a[j] == 1)
        p |= (b[j] & 1), q |= (b[j] & 1);
      else if (a[j] == 2)
        p &= (b[j] & 1), q &= (b[j] & 1);
      else
        p ^= (b[j] & 1), q ^= (b[j] & 1);
      b[j] >>= 1;
    }
    if (p == 0 && q == 0) {
    } else if (p == 0 && q == 1) {
      a1 += 1 << i;
    } else if (p == 1 && q == 0) {
      a1 += 1 << i;
      a3 += 1 << i;
    } else {
      a1 += 1 << i;
      a2 += 1 << i;
    }
  }
  cout << 3 << endl;
  cout << '&' << ' ' << a1 << endl;
  cout << '|' << ' ' << a2 << endl;
  cout << '^' << ' ' << a3 << endl;
}
