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;
int gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int ind[4];
  ind[0] = 1;
  ind[1] = 1;
  ind[2] = 0;
  ind[3] = 0;
  int modi[4];
  modi[0] = 0;
  modi[1] = 1;
  modi[2] = 1;
  modi[3] = 1;
  int op[2][2];
  op[0][0] = 0;
  op[0][1] = 1;
  op[1][0] = 2;
  op[1][1] = 3;
  char com[n];
  int c[n];
  for (int i = 0; i < n; i++) {
    cin >> com[i] >> c[i];
  }
  int CC[] = {0, 0, 0, 0};
  for (int b = 0; b < 10; b++) {
    int r0 = 0 << b;
    int r1 = 1 << b;
    for (int i = 0; i < n; i++) {
      if (com[i] == '|') {
        r0 = r0 | c[i];
        r1 = r1 | c[i];
      } else if (com[i] == '^') {
        r0 = r0 ^ c[i];
        r1 = r1 ^ c[i];
      } else if (com[i] == '&') {
        r0 = r0 & c[i];
        r1 = r1 & c[i];
      }
    }
    r0 = r0 & (1 << b) ? 1 : 0;
    r1 = r1 & (1 << b) ? 1 : 0;
    int ope = op[r0][r1];
    for (int i = 0; i < 4; i++) {
      if (i == ope) {
        CC[i] += (modi[i] << b);
      } else {
        CC[i] += (ind[i] << b);
      }
    }
  }
  char oper[] = {'&', '&', '^', '|'};
  cout << 4 << endl;
  for (int i = 0; i < 4; i++) {
    cout << oper[i] << " " << CC[i] << endl;
  }
}