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 int N = 3e5 + 10;
long long test, n, m, k, x, y, z, l, ans, r[N], g[N], b[N], mod = 1e9 + 7;
string s, t;
int T, nr, ng, nb;
void solve() {
  cin >> n;
  long long k = 0, l = 1023;
  for (long long i = 1; i <= n; i++) {
    char ch;
    int m;
    cin >> ch >> m;
    if (ch == '|') k |= m, l |= m;
    if (ch == '&') k &= m, l &= m;
    if (ch == '^') k ^= m, l ^= m;
  }
  cout << 3 << "\n";
  cout << '|' << " " << (k & l) << "\n";
  cout << '&' << " " << (k | l) << "\n";
  cout << '^' << " " << (k & (1023 ^ l)) << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}