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 ll = long long;
using pll = pair<ll, ll>;
using dd = long double;
namespace {
static bool constexpr dbg = 0;
ll constexpr N = 11;
ll n;
struct T {
  ll v, f;
};
T t[N + 9];
void init() { cin >> n; }
void solve() {
  for (ll _ = 1, ii_ = (n); _ <= ii_; ++_) {
    char ty;
    ll v;
    cin >> ty >> v;
    for (ll i = 0, ii_ = (10); i < ii_; ++i) {
      ll bit = v & 1;
      v /= 2;
      if (ty == '|' && bit == 1) {
        t[i].v = 1;
        t[i].f = 0;
      } else if (ty == '&' && bit == 0) {
        t[i].v = -1;
        t[i].f = 0;
      } else if (ty == '^' && bit == 1) {
        ++t[i].f;
      }
    }
  }
  cout << "3\n";
  ll v = 0;
  for (ll i = 0, ii_ = (10); i < ii_; ++i)
    if (t[i].v == 1) v += (1 << i);
  cout << "| " << v << '\n';
  v = 1023;
  for (ll i = 0, ii_ = (10); i < ii_; ++i)
    if (t[i].v == -1) v -= (1 << i);
  cout << "& " << v << '\n';
  v = 0;
  for (ll i = 0, ii_ = (10); i < ii_; ++i)
    if (t[i].f & 1) v += (1 << i);
  cout << "^ " << v << '\n';
}
}  // namespace
int main() {
  if (!dbg) {
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  cout << fixed << setprecision(20);
  init();
  solve();
  return 0;
}