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 arr[100000];
long long arr1[600000][10];
void fill(long long* arr, long long n) {
  for (long long i = 0; i < n; i++) {
    long long num = arr[i];
    for (long long j = 9; j >= 0; j--) {
      arr1[i][j] = num % 2;
      num = num / 2;
    }
  }
}
long long getnum(long long arrr[]) {
  long long num = arrr[0];
  for (long long i = 1; i < 10; i++) {
    cout << arrr[i] << endl;
    num = num * 2 + arrr[i];
  }
  return num;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  char s[n][3];
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> s[i] >> a[i];
  fill(a, n);
  long long arr[10][2];
  for (long long i = 0; i < 10; i++) {
    arr[i][0] = 0;
    arr[i][1] = 1;
  }
  for (long long i = 0; i < 10; i++) {
    for (long long j = 0; j < n; j++) {
      if (s[j][0] == '^')
        arr[i][0] ^= arr1[j][i];
      else if (s[j][0] == '&')
        arr[i][0] &= arr1[j][i];
      else if (s[j][0] == '|')
        arr[i][0] |= arr1[j][i];
      if (s[j][0] == '^')
        arr[i][1] ^= arr1[j][i];
      else if (s[j][0] == '&')
        arr[i][1] &= arr1[j][i];
      if (s[j][0] == '|') arr[i][1] |= arr1[j][i];
    }
  }
  vector<long long> inv, sam, zer, one;
  for (long long i = 0; i < 10; i++) {
    if (arr[i][0] == 0 && arr[i][1] == 0)
      zer.push_back(i);
    else if (arr[i][0] == 1 && arr[i][1] == 1)
      one.push_back(i);
    else if (arr[i][0] == 0 && arr[i][1] == 1)
      sam.push_back(i);
    else
      inv.push_back(i);
  }
  cout << '3' << endl;
  cout << '&' << ' ';
  long long b[10];
  for (long long i = 0; i < 10; i++) b[i] = 1;
  for (long long i = 0; i < zer.size(); i++) {
    b[zer[i]] = 0;
  }
  long long num = b[0];
  for (long long i = 1; i < 10; i++) {
    num = num * 2 + b[i];
  }
  cout << num << endl;
  cout << '|' << ' ';
  for (long long i = 0; i < 10; i++) b[i] = 0;
  for (long long i = 0; i < one.size(); i++) b[one[i]] = 1;
  num = b[0];
  for (long long i = 1; i < 10; i++) {
    num = num * 2 + b[i];
  }
  cout << num << endl;
  cout << '^' << ' ';
  for (long long i = 0; i < 10; i++) b[i] = 0;
  for (long long i = 0; i < inv.size(); i++) b[inv[i]] = 1;
  num = b[0];
  for (long long i = 1; i < 10; i++) {
    num = num * 2 + b[i];
  }
  cout << num << endl;
}