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 N = 500020;
char t[N];
int a[N];
int inv[N], ans[N];
int main(void) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf(" %c %d", t + i, a + i);
  }
  memset(ans, -1, sizeof ans);
  for (int bit = 0; bit <= 9; bit++) {
    bool inverte = false;
    int x = -1;
    for (int i = 0; i < n; i++) {
      int y = ((1 << bit) & a[i]) ? 1 : 0;
      if (!y and t[i] == '|') continue;
      if (!y and t[i] == '^') continue;
      if (y and t[i] == '&') continue;
      if (t[i] == '|') {
        x = 1;
        inverte = false;
      }
      if (t[i] == '^') {
        if (x == -1)
          inverte = !inverte;
        else if (x == 1)
          x = 0;
        else
          x = 1;
      }
      if (t[i] == '&') {
        x = 0;
        inverte = false;
      }
    }
    if (x == -1)
      inv[bit] = inverte;
    else
      ans[bit] = x;
  }
  int xorr = 0, orr = 0, andd = 1023;
  for (int bit = 0; bit <= 9; bit++) {
    if (ans[bit] == -1) {
      if (inv[bit]) xorr |= (1 << bit);
    } else if (ans[bit] == 1) {
      orr |= (1 << bit);
    } else {
      andd ^= (1 << bit);
    }
  }
  printf("3\n| %d\n^ %d\n& %d\n", orr, xorr, andd);
  return 0;
}
