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;
const int K = 10;
const int N = 500500;
int n;
int a[N];
char b[N];
bool Res[K][2];
int f(int i, int x, int bit) {
  if (i == n) return x;
  bool cur = (a[i] & (1 << bit));
  if (b[i] == '&') return f(i + 1, x & cur, bit);
  if (b[i] == '|') return f(i + 1, x | cur, bit);
  if (b[i] == '^') return f(i + 1, x ^ cur, bit);
  assert(0);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf(" %c%d", b + i, a + i);
  for (int bit = 0; bit < K; bit++) {
    Res[bit][0] = f(0, 0, bit);
    Res[bit][1] = f(0, 1, bit);
  }
  int OrMask = 0;
  int AndMask = (1 << K) - 1;
  int XorMask = 0;
  for (int bit = 0; bit < K; bit++) {
    if (Res[bit][0] != Res[bit][1]) {
      if (Res[bit][0] == 1) XorMask |= (1 << bit);
    } else {
      if (Res[bit][0] == 0)
        AndMask ^= (1 << bit);
      else
        OrMask |= (1 << bit);
    }
  }
  puts("3");
  cout << "| " << OrMask << endl;
  cout << "& " << AndMask << endl;
  cout << "^ " << XorMask << endl;
}
