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;
long long n, k;
int zym[10];
int xooor[10];
bool oneth[10];
bool yes[10];
int main() {
  while (cin >> n) {
    char ch;
    int x;
    memset(oneth, true, sizeof oneth);
    for (int i = 1; i <= n; i++) {
      scanf(" %c %d", &ch, &x);
      if (ch == '|') {
        for (int i = 0; i < 10; i++) {
          if ((1 << i) & x) {
            yes[i] = true;
            zym[i] = 1;
          }
        }
      } else if (ch == '^') {
        for (int i = 0; i < 10; i++) {
          if (yes[i] == true) {
            if ((1 << i) & x)
              zym[i] ^= 1;
            else
              zym[i] ^= 0;
          } else {
            if (oneth[i]) {
              xooor[i] = (x >> i) & 1;
              oneth[i] = false;
            } else
              xooor[i] ^= (x >> i) & 1;
          }
        }
      } else {
        for (int i = 0; i < 10; i++) {
          if (!((1 << i) & x)) {
            yes[i] = true;
            zym[i] = 0;
          }
        }
      }
    }
    int a = 0, b = (1 << 10) - 1, c = 0;
    for (int i = 0; i < 10; i++) {
      if (yes[i]) {
        if (zym[i] == 1)
          a |= 1 << i;
        else {
          int tmp = 0;
          for (int j = 0; j < 10; j++) {
            if (((1 << j) & b) && i != j) tmp |= (1 << j);
          }
          b = tmp;
        }
      } else {
        if (xooor[i] == 1) c |= (1 << i);
      }
    }
    cout << 3 << endl;
    cout << "| " << a << endl;
    cout << "& " << b << endl;
    cout << "^ " << c << endl;
  }
  return 0;
}
