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 power(int base, int pow, int mo) {
  int res = 1;
  while (pow) {
    if (pow % 2 != 0) res = (res * base) % mo;
    base = (base * base) % mo;
    pow /= 2;
  }
  return res;
}
int nCrModpDP(int n, int r, int p) {
  int C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
int nCrModpLucas(int n, int r, int p) {
  if (r == 0) return 1;
  int ni = n % p, ri = r % p;
  return (nCrModpLucas(n / p, r / p, p) * nCrModpDP(ni, ri, p)) % p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t, i, j, x, m, n, k;
  t = 1;
  while (t--) {
    cin >> n;
    int a = 0, b = 1023;
    for (i = 0; i < n; i++) {
      char c;
      cin >> c >> x;
      if (c == '|')
        a |= x, b |= x;
      else if (c == '&')
        a &= x, b &= x;
      else
        a ^= x, b ^= x;
    }
    cout << 3 << "\n";
    cout << "^ " << (a & ~b & 1023) << "\n";
    cout << "| " << (a & b) << "\n";
    cout << "& " << (a | b) << "\n";
  }
  return 0;
}