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 = (1 << 20);
int n, m, val, g[66][2], to[2], f[2];
char s[N];
int main() {
  scanf("%d", &n);
  for (int i = 0, iE = 9; i <= iE; i++) g[i][0] = 0, g[i][1] = 1;
  while (n--) {
    scanf("%s", s + 1);
    scanf("%d", &val);
    for (int i = 9, iE = 0; i >= iE; i--) {
      f[0] = g[i][0], f[1] = g[i][1];
      if (val >> i & 1) {
        if (s[1] == '|') to[0] = 1, to[1] = 1;
        if (s[1] == '&') to[0] = 0, to[1] = 1;
        if (s[1] == '^') to[0] = 1, to[1] = 0;
      } else {
        if (s[1] == '|') to[0] = 0, to[1] = 1;
        if (s[1] == '&') to[0] = 0, to[1] = 0;
        if (s[1] == '^') to[0] = 0, to[1] = 1;
      }
      g[i][0] = to[g[i][0]], g[i][1] = to[g[i][1]];
    }
  }
  int val1 = 0, val2 = 0, val3 = 0;
  for (int i = 0, iE = 9; i <= iE; i++) {
    if (g[i][0] == 1 && g[i][1] == 1) val1 |= (1 << i);
    if (!(g[i][0] == 0 && g[i][1] == 0)) val2 |= (1 << i);
    if (g[i][0] == 1 && g[i][1] == 0) val3 |= (1 << i);
  }
  puts("3");
  printf("| %d\n", val1);
  printf("& %d\n", val2);
  printf("^ %d\n", val3);
  return 0;
}