#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
long long N;
bool checkbit(long long n, long long p) { return n & (1ll << p); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  vector<long long> a(10);
  for (long long i = 0; i < N; i++) {
    char cm;
    long long r;
    cin >> cm >> r;
    if (cm == '&') {
      for (long long p = 0; p < 10; p++) {
        if (!checkbit(r, p)) {
          a[p] = 1;
        }
      }
    } else if (cm == '|') {
      for (long long p = 0; p < 10; p++) {
        if (checkbit(r, p)) {
          a[p] = 2;
        }
      }
    } else if (cm == '^') {
      for (long long p = 0; p < 10; p++) {
        if (checkbit(r, p)) {
          if (a[p] == 1) {
            a[p] = 2;
          } else if (a[p] == 2) {
            a[p] = 1;
          } else if (a[p] == 3) {
            a[p] = 0;
          } else {
            a[p] = 3;
          }
        }
      }
    }
  }
  bitset<10> n1, n2, n3;
  n1.set();
  for (long long i = 0; i < 10; i++) {
    if (a[i] == 1) {
      n1.set(i, 0);
    } else if (a[i] == 2) {
      n2.set(i, 1);
    } else if (a[i] == 3) {
      n3.set(i, 1);
    }
  }
  cout << 3 << endl;
  cout << "^ " << n3.to_ulong() << endl;
  cout << "| " << n2.to_ulong() << endl;
  cout << "& " << n1.to_ulong() << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
long long N;
bool checkbit(long long n, long long p) { return n & (1ll << p); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  vector<long long> a(10);
  for (long long i = 0; i < N; i++) {
    char cm;
    long long r;
    cin >> cm >> r;
    if (cm == '&') {
      for (long long p = 0; p < 10; p++) {
        if (!checkbit(r, p)) {
          a[p] = 1;
        }
      }
    } else if (cm == '|') {
      for (long long p = 0; p < 10; p++) {
        if (checkbit(r, p)) {
          a[p] = 2;
        }
      }
    } else if (cm == '^') {
      for (long long p = 0; p < 10; p++) {
        if (checkbit(r, p)) {
          if (a[p] == 1) {
            a[p] = 2;
          } else if (a[p] == 2) {
            a[p] = 1;
          } else if (a[p] == 3) {
            a[p] = 0;
          } else {
            a[p] = 3;
          }
        }
      }
    }
  }
  bitset<10> n1, n2, n3;
  n1.set();
  for (long long i = 0; i < 10; i++) {
    if (a[i] == 1) {
      n1.set(i, 0);
    } else if (a[i] == 2) {
      n2.set(i, 1);
    } else if (a[i] == 3) {
      n3.set(i, 1);
    }
  }
  cout << 3 << endl;
  cout << "^ " << n3.to_ulong() << endl;
  cout << "| " << n2.to_ulong() << endl;
  cout << "& " << n1.to_ulong() << endl;
  return 0;
}

```