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;
const int N = 5000010;
int n;
bool a[20], b[20];
char ch[20];
string a1, a2, a3;
int main() {
  for (int i = 0; i < 10; ++i) {
    a[i] = 0;
    b[i] = 1;
  }
  bool kk = 0, e = 2;
  cin >> n;
  while (n--) {
    int num;
    scanf("%s%d", ch, &num);
    for (int i = 9, j = 0; i >= 0; --i, ++j) {
      bool ss = (num >> i) & 1;
      if (ch[0] == '&') {
        b[i] &= ss;
        a[i] &= ss;
      } else if (ch[0] == '|') {
        b[i] |= ss;
        a[i] |= ss;
      } else
        b[i] ^= ss, a[i] ^= ss;
    }
  }
  for (int i = 0; i < 10; ++i) {
    if (b[i] == 0 && a[i] == 0) {
      a1.push_back('1');
      a2.push_back('1');
      a3.push_back('1');
    } else if (b[i] == 1 && a[i] == 1) {
      a1.push_back('1');
      a2.push_back('1');
      a3.push_back('0');
    } else if (b[i] == 1 && a[i] == 0) {
      a1.push_back('1');
      a2.push_back('0');
      a3.push_back('0');
    } else {
      a1.push_back('1');
      a2.push_back('0');
      a3.push_back('1');
    }
  }
  cout << 3 << endl;
  int num1 = 0, num2 = 0, num3 = 0, ba = 1;
  for (int i = 0; i < 10; ++i) {
    num1 += (a1[i] - '0') * ba;
    num2 += (a2[i] - '0') * ba;
    num3 += (a3[i] - '0') * ba;
    ba *= 2;
  }
  cout << "& " << num1 << endl;
  cout << "| " << num2 << endl;
  cout << "^ " << num3 << endl;
}