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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
int a, b, c, d, n, m, k;
int zz, oo;
char str[3];
const char* cc = "^|&";
inline void upd(int& first, int o, int v) {
  if (o == 0)
    first ^= v;
  else if (o == 1)
    first |= v;
  else
    first &= v;
}
int main() {
  scanf("%d", &n);
  zz = 0;
  oo = 1023;
  for (int _n((n)-1), i(0); i <= _n; i++) {
    scanf("%s%d", str, &a);
    if (str[0] == '^') {
      zz ^= a;
      oo ^= a;
    } else if (str[0] == '|') {
      zz |= a;
      oo |= a;
    } else {
      zz &= a;
      oo &= a;
    }
  }
  for (int _n((3) - 1), s1(0); s1 <= _n; s1++) {
    for (int _n((1024) - 1), i(0); i <= _n; i++) {
      for (int _n((3) - 1), s2(0); s2 <= _n; s2++) {
        for (int _n((1024) - 1), j(0); j <= _n; j++) {
          int cz = 0, co = 1023;
          upd(cz, s1, i);
          upd(co, s1, i);
          upd(cz, s2, j);
          upd(co, s2, j);
          if (cz == zz && co == oo) {
            printf("2\n");
            printf("%c %d\n", cc[s1], i);
            printf("%c %d\n", cc[s2], j);
            exit(0);
          }
        }
      }
    }
  }
}
