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 <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937_64 rng(61378913);
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
long long n, m, k, q, l, r, x, y, z;
const long long template_array_size = 1e6 + 8432;
long long a[template_array_size];
long long b[template_array_size];
long long c[template_array_size];
string second, t;
long long ans = 0;
void solve(int tc = 0) {
  bool dead[10] = {0};
  long long v = 0;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    char c;
    cin >> c >> x;
    if (c == '|') {
      for (long long j = 0; j < 10; j++) {
        if (x & (1 << j)) dead[j] = 1;
      }
      v |= x;
    } else if (c == '&') {
      for (long long j = 0; j < 10; j++) {
        if (!(x & (1 << j))) dead[j] = 1;
      }
      v &= x;
    } else {
      v ^= x;
    }
  }
  long long ov = 0;
  for (long long i = 0; i < 10; i++) {
    if (!dead[i]) ov += (1 << i);
  }
  cout << 2 << '\n';
  cout << "& " << ov << '\n';
  cout << "^ " << v << '\n';
}
int main() {
  { ios_base::sync_with_stdio(false); }
  { cin.tie(NULL); }
  cout << setprecision(12) << fixed;
  int tc = 1;
  for (int t = 0; t < tc; t++) solve(t);
}