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.
#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long N = 2002;
long long dr[4] = {0, 0, 1, -1};
long long dc[4] = {1, -1, 0, 0};
long long ddr[8] = {0, 0, 1, -1, 1, -1, 1, -1};
long long ddc[8] = {1, -1, 0, 0, 1, -1, -1, 1};
long long ddr1[8] = {1, 1, -1, -1, 2, 2, -2, -2};
long long ddc1[8] = {2, -2, 2, -2, 1, -1, 1, -1};
int32_t main() {
  fast();
  long long n, x, x1 = 0, x2 = 1023;
  long long y1 = 0, y2 = 0, y3 = 0;
  y1 = x2;
  char ch;
  cin >> n;
  while (n--) {
    cin >> ch >> x;
    if (ch == '&') {
      x1 &= x;
      x2 &= x;
    } else if (ch == '^') {
      x1 ^= x;
      x2 ^= x;
    } else {
      x1 |= x;
      x2 |= x;
    }
  }
  for (long long i = 0; i < 10; i++) {
    bool f1, f2;
    long long t1 = (1 << i);
    f1 = (x1 & t1);
    f2 = (x2 & t1);
    if (f1 == 1 && f2 == 0)
      y3 |= t1;
    else if (f1 == 1 && f2 == 1)
      y2 |= t1;
    else if (f1 == 0 && f2 == 0)
      y1 ^= t1;
  }
  cout << "3\n& " << y1 << "\n| " << y2 << "\n^ " << y3;
}
