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;
inline void read(int &x) {
  char ch = getchar();
  x = 0;
  int e = 1;
  for (; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') e = -1, ch = getchar();
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - 48, ch = getchar())
    ;
  x *= e;
}
inline void read(long long &x) {
  char ch = getchar();
  x = 0;
  int e = 1;
  for (; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') e = -1, ch = getchar();
  for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - 48, ch = getchar())
    ;
  x *= e;
}
inline void write(int data) {
  if (data < 0) data = -data, putchar('-');
  if (data > 9) write(data / 10);
  putchar(data % 10 + 48);
}
inline void write(long long data) {
  if (data < 0) data = -data, putchar('-');
  if (data > 9) write(data / 10);
  putchar(data % 10 + 48);
}
int a0[10], a1[10];
int n;
int opera(int x, char ch, int y) {
  if (ch == '|') return x | y;
  if (ch == '&') return x & y;
  if (ch == '^') return x ^ y;
}
int main() {
  read(n);
  for (int i = 0; i <= 9; ++i) a0[i] = 0, a1[i] = 1;
  char ch;
  int val;
  for (int i = 1; i <= n; ++i) {
    ch = getchar(), read(val);
    for (int j = 0; j <= 9; ++j) {
      int Val = ((1 << j) & val) > 0;
      a0[j] = opera(a0[j], ch, Val), a1[j] = opera(a1[j], ch, Val);
    }
  }
  int valor = 0, valand = (1 << 10) - 1, valxor = 0;
  for (int i = 0; i <= 9; ++i) {
    int Val = 1 << i;
    if (a1[i] == 1 && a0[i] == 1) valor |= Val;
    if (a1[i] == 0 && a0[i] == 0) valand ^= Val;
    if (a1[i] == 0 && a0[i] == 1) valxor |= Val;
  }
  puts("3");
  cout << "| " << valor << endl;
  cout << "& " << valand << endl;
  cout << "^ " << valxor << endl;
}
