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 MAXN = (int)5e5 + 5;
char a[MAXN];
int b[MAXN];
int cal(const int n, const int p, const int u) {
  int cur(0);
  if (u == 1) cur |= (1 << p);
  for (int i(1); i <= n; ++i) {
    if (a[i] == '&')
      cur &= b[i];
    else if (a[i] == '|')
      cur |= b[i];
    else
      cur ^= b[i];
  }
  if (u == 0) return cur;
  int res = (cur >> p) & 1;
  return res;
}
void solve(const int n) {
  int x(0), y(0), z((1 << 10) - 1);
  int tmp = cal(n, 0, 0);
  for (int p(0); p < 10; ++p) {
    int change0 = (tmp >> p) & 1;
    int change1 = cal(n, p, 1);
    if (change0 == 1 && change1 == 1)
      y |= (1 << p);
    else if (change0 == 0 && change1 == 0)
      z -= (1 << p);
    else if (change0 == 1 && change1 == 0)
      x |= (1 << p);
  }
  puts("3");
  printf("^ %d\n", x);
  printf("| %d\n", y);
  printf("& %d\n", z);
}
int main() {
  int n;
  scanf("%d", &n);
  char tmp[5];
  for (int i(1); i <= n; ++i) {
    scanf("%s %d", tmp, b + i);
    a[i] = tmp[0];
  }
  solve(n);
  return 0;
}