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 long long oo = 1000000000000000000;
const int N = 100005;
int f(int x) {
  x |= 3;
  x ^= 2;
  return x;
}
int f2(int x) { return (x & 1021) | 1; }
int v[10];
int main() {
  int n;
  scanf("%d", &n);
  char c;
  int x;
  for (int i = 0; i < int(n); i++) {
    scanf(" %c %d", &c, &x);
    if (c == '|') {
      for (int i = 0; i < int(10); i++)
        if (x & (1 << i)) {
          v[i] = 3;
        }
    } else if (c == '&') {
      for (int i = 0; i < int(10); i++) {
        if (x & (1 << i))
          ;
        else {
          v[i] = 2;
        }
      }
    } else if (c == '^') {
      for (int i = 0; i < int(10); i++) {
        if (x & (1 << i)) {
          if (!v[i])
            v[i] = 1;
          else if (v[i] == 1)
            v[i] = 0;
          else if (v[i] == 2)
            v[i] = 3;
          else if (v[i] == 3)
            v[i] = 2;
        }
      }
    }
  }
  vector<pair<char, int> > ans;
  x = 0;
  for (int i = 0; i < int(10); i++)
    if (v[i] == 1) x |= 1 << i;
  if (x) ans.emplace_back('^', x);
  x = 1023;
  for (int i = 0; i < int(10); i++)
    if (v[i] == 2) x &= ~(1 << i);
  if (x != 1023) ans.emplace_back('&', x);
  x = 0;
  for (int i = 0; i < int(10); i++)
    if (v[i] == 3) x |= 1 << i;
  if (x) ans.emplace_back('|', x);
  printf("%d\n", int(ans.size()));
  for (int i = 0; i < int(ans.size()); i++)
    printf("%c %d\n", ans[i].first, ans[i].second);
  return 0;
}