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.
#include <bits/stdc++.h>
using namespace std;
struct S {
  int a, b;
  S() {}
  S(int _a, int _b) {
    a = _a;
    b = _b;
  }
  const bool operator<(const S &o) const { return a < o.a; }
};
priority_queue<int, vector<int>, greater<int>> mpq;
const int SPACE = 0, NL = 1;
string exm;
inline void showAll(vector<int> &v, int sep) {
  for (int &here : v) printf("%d%c", here, (sep) ? '\n' : ' ');
}
inline void exf(void) {
  cout << exm << "\n";
  exit(0);
}
inline vector<int> int_seperation(int N, int d = 10) {
  vector<int> v;
  while (N) {
    v.push_back(N % d);
    N /= d;
  }
  reverse(v.begin(), v.end());
  return v;
}
int main() {
  int n;
  scanf("%d", &n);
  int _and = 0, _or = 0, _xor = 0;
  int _0 = 0, _1023 = 1023;
  while (n--) {
    char op;
    int k;
    scanf(" %c %d", &op, &k);
    if (op == '&') {
      _0 &= k;
      _1023 &= k;
    } else if (op == '|') {
      _0 |= k;
      _1023 |= k;
    } else {
      _0 ^= k;
      _1023 ^= k;
    }
  }
  for (int i = 0; i < 10; i++) {
    int b0 = _0 & (1 << i), b1 = _1023 & (1 << i);
    if (b0 && b1) {
      _or += (1 << i);
      _and += (1 << i);
    } else if (!b0 && b1) {
      _and += (1 << i);
    } else if (b0 && !b1) {
      _xor += (1 << i);
      _and += (1 << i);
    }
  }
  printf("3\n");
  printf("& %d\n", _and);
  printf("^ %d\n", _xor);
  printf("| %d\n", _or);
  return 0;
}
