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;
template <class C>
void mini(C &a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C &a4, C b4) {
  a4 = max(a4, b4);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
bool bit(long long a, long long b) { return a & (1 << b); }
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  if (0) cout << fixed << setprecision(6);
  long long n;
  cin >> n;
  long long a = 1023;
  long long b = 0;
  for (long long i = (0); i <= ((long long)(n)-1); i++) {
    char c;
    long long x;
    cin >> c >> x;
    switch (c) {
      case '&':
        a &= x;
        b &= x;
        break;
      case '|':
        a |= x;
        b |= x;
        break;
      case '^':
        a ^= x;
        b ^= x;
    }
    (c, x);
  }
  (a, b);
  long long a1 = 0;
  long long a2 = 1023;
  long long a3 = 0;
  for (long long i = (0); i <= ((long long)(10) - 1); i++) {
    if (bit(a, i) && bit(b, i)) {
      a1 |= 1 << i;
    } else if (!bit(a, i) && !bit(b, i)) {
      a2 ^= 1 << i;
    } else if (!bit(a, i) && bit(b, i)) {
      a3 |= 1 << i;
    }
  }
  cout << 3 << endl;
  cout << "| " << a1 << endl;
  cout << "& " << a2 << endl;
  cout << "^ " << a3 << endl;
  assert((((1023 | a1) & a2) ^ a3) == a);
  assert((((0 | a1) & a2) ^ a3) == b);
}