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;
using INT = long long;
using pci = pair<char, int>;
const int N = 1023;
pci p[500100];
char s[] = {
    '&',
    '|',
    '^',
};
int ans[20][2];
int val[500100];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    char c;
    int first;
    scanf(" %c %d", &c, &first);
    p[i] = pci(c, first);
  }
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 2; j++) {
      ans[i][j] = j;
      for (int k = 1; k <= n; k++) {
        int tmp = p[k].second;
        int op = (tmp & (1 << i));
        if (op) op = 1;
        if (p[k].first == s[0]) ans[i][j] = (ans[i][j] & op);
        if (p[k].first == s[1]) ans[i][j] = (ans[i][j] | op);
        if (p[k].first == s[2]) ans[i][j] ^= op;
      }
    }
  }
  int first = 0, second = 0;
  for (int i = 0; i < 10; i++) {
    if (ans[i][0] == ans[i][1]) {
      first += (1 << i);
      second += (ans[i][0] ^ 1) * (1 << i);
    } else {
      second += (ans[i][0] ^ 0) * (1 << i);
    }
  }
  cout << 2 << endl;
  cout << s[1] << " " << first << endl;
  cout << s[2] << " " << second << endl;
  return 0;
}