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;
long long int cons;
long long int check(long long int a) {
  if (a >= cons || a <= -cons) a %= cons;
  return a;
}
long long int check2(long long int a) {
  if (a > 0) return a;
  long long int b = a / cons;
  a -= b * cons;
  if (a < 0) a += cons;
  return a;
}
long long int GCD(long long int a, long long int b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
long long int exp(long long int a, long long int n) {
  if (n == 0) return 1;
  if (n == 1) return check(a);
  long long int b = exp(a, n / 2);
  if (n % 2 == 0) return check(b * b);
  return check(b * check(b * a));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cons = 1e9 + 7;
  int n;
  cin >> n;
  vector<int> arr;
  vector<char> arr2;
  arr2.resize(n);
  arr.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> arr2[i];
    cin >> arr[i];
  }
  vector<int> dp;
  dp.resize(10);
  for (int i = 0; i < 10; i++) dp[i] = 2;
  for (int i = 0; i < n; i++) {
    if (arr2[i] == '|') {
      for (int j = 0; j < 10; j++) {
        if (arr[i] & (1 << j)) dp[j] = 1;
      }
    } else if (arr2[i] == '&') {
      for (int j = 0; j < 10; j++) {
        if (!(arr[i] & (1 << j))) dp[j] = 0;
      }
    } else {
      for (int j = 0; j < 10; j++) {
        if ((arr[i] & (1 << j))) {
          if (dp[j] < 2)
            dp[j] = 1 - dp[j];
          else
            dp[j] = 5 - dp[j];
        }
      }
    }
  }
  cout << 3 << endl;
  cout << "& ";
  int temp = 1023;
  for (int j = 0; j < 10; j++)
    if (dp[j] == 0) temp -= (1 << j);
  cout << temp << endl;
  cout << "| ";
  temp = 0;
  for (int i = 0; i < 10; i++)
    if (dp[i] == 1) temp += (1 << i);
  cout << temp << endl;
  cout << "^ ";
  temp = 0;
  for (int i = 0; i < 10; i++)
    if (dp[i] == 3) temp += (1 << i);
  cout << temp;
}