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>
const int N = 15;
using namespace std;
int n;
int g[N], hlp[N];
void READ() {
  scanf("%d", &n);
  for (int i = 0; i <= 9; i++) g[i] = -1, hlp[i] = 0;
}
void SOLVE() {
  char pd;
  int num;
  scanf("\n%c%d", &pd, &num);
  if (pd == '|') {
    for (int i = 9; i >= 0; i--) {
      int now = ((num >> i) & 1);
      if (now) g[i] = 1;
    }
  }
  if (pd == '&') {
    for (int i = 9; i >= 0; i--) {
      int now = ((num >> i) & 1);
      if (!now) g[i] = 0;
    }
  }
  if (pd == '^') {
    for (int i = 9; i >= 0; i--) {
      int now = ((num >> i) & 1);
      if (now == 0) continue;
      if (g[i] == -1)
        hlp[i] ^= 1;
      else
        g[i] ^= now;
    }
  }
}
void GETANS() {
  printf("3\n");
  int num = 0;
  for (int i = 0; i <= 9; i++)
    if (g[i] == -1) num |= hlp[i] * (1 << i);
  printf("^ %d\n", num);
  num = 1023;
  for (int i = 0; i <= 9; i++)
    if (g[i] == 0) num ^= (1 << i);
  printf("& %d\n", num);
  num = 0;
  for (int i = 0; i <= 9; i++)
    if (g[i] == 1) num |= (1 << i);
  printf("| %d\n", num);
}
int main() {
  READ();
  for (int i = 1; i <= n; i++) SOLVE();
  GETANS();
  return 0;
}
