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 long long M = 1e9 + 7;
const long long N = 2000 + 10;
const long long IM = 1e15 + 37;
const long double PI = 3.1415926535897932384;
inline void PP(long long n) { cout << n << " "; }
void PV(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
signed main() {
  op();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  bitset<10> a, b(1023);
  long long n;
  cin >> n;
  long long ok = 1;
  for (long long i = 0; i < n; i++) {
    char c;
    long long x;
    cin >> c >> x;
    bitset<10> d(x);
    if (c == '|') {
      a = a | d;
      b = b | d;
      ok = ok | x;
    } else if (c == '&') {
      a = a & d;
      b = b & d;
      ok = ok & x;
    } else {
      a = a ^ d;
      b = b ^ d;
      ok = ok ^ x;
    }
  }
  long long x = 0, y = 0, u = 1, z = 0, zz = 1023;
  for (long long i = 0; i < 10; i++) {
    if (a[i] == 0 && b[i] == 1) {
      u = u * 2;
      continue;
    } else if (a[i] && b[i])
      x = x | u;
    else if (a[i] == 1 && b[i] == 0) {
      z = z | u;
    }
    u = u * 2;
  }
  u = 1;
  for (long long i = 0; i < 10; i++) {
    if (a[i] == 0 && b[i] == 0) {
      u = u * 2;
      continue;
    }
    y = y | u;
    u = u * 2;
  }
  zz = 0;
  cout << 4 << "\n";
  cout << "^ " << z << "\n";
  cout << "^ " << zz << "\n";
  cout << "& " << y << "\n";
  cout << "| " << x << "\n";
  return 0;
}