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;
vector<long long int> primes;
void sieve(long long int n) {
  vector<bool> check(n + 1, true);
  check[0] = false;
  check[1] = false;
  for (long long int i = 2; i <= n; ++i) {
    if (check[i]) {
      if (i * i * 1ll <= n)
        for (long long int j = i * i; j <= n; j += i) check[j] = false;
    }
  }
  for (long long int i = 0; i < n + 1; ++i) {
    if (check[i]) primes.push_back(i);
  }
}
vector<int> v[105];
bool visited[105];
int maxi = 0, budget = 0;
;
vector<int> stackprototype;
void dfs(int source) {
  visited[source] = true;
  for (int i = 0; i < v[source].size(); ++i) {
    if (visited[v[source][i]] == false) {
      dfs(v[source][i]);
    }
  }
  if (v[source].size() == 1) {
    stackprototype.push_back(source);
  }
  return;
}
bool isPrime(long long int n) {
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  int test = 1;
  while (test--) {
    int n, x = 0, y = 1023, w;
    char ch;
    for (cin >> n; n--;) {
      cin >> ch >> w;
      if (ch == '^') {
        x ^= w;
        y ^= w;
      } else if (ch == '&') {
        x &= w;
        y &= w;
      } else {
        x |= w;
        y |= w;
      }
    }
    cout << "3" << endl;
    cout << "^ " << (x & ~y & 1023) << "\n& " << (x | y) << "\n| " << (x & y);
  }
  return 0;
}