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;
typedef long double ld;
const ld eps = 1e-8;
const ld tau = 2 * acosl(-1);
const int inf = 1e9 + 99;
const long long linf = 1e18 + 99;
const int P = 1e9 + 7;
bool flip[10], yset[10], nset[10];
int32_t main() {
  int n;
  scanf("%d", &n);
  for (; n--;) {
    static char buf[4];
    int x;
    scanf("%s%d", buf, &x);
    if (*buf == '|') {
      for (int i = 0; i < 10; i++)
        if (x & (1 << i)) yset[i] = 1, nset[i] = 0;
    } else if (*buf == '&') {
      for (int i = 0; i < 10; i++)
        if (!(x & (1 << i))) nset[i] = 1, yset[i] = 0;
    } else if (*buf == '^') {
      for (int i = 0; i < 10; i++)
        if (x & (1 << i)) {
          swap(yset[i], nset[i]);
          flip[i] ^= 1;
        }
    } else
      assert(0);
  }
  int xx = 0, oo = 0, aa = 1023;
  for (int i = 0; i < 10; i++)
    if (flip[i]) xx |= (1 << i);
  for (int i = 0; i < 10; i++)
    if (yset[i]) oo |= (1 << i);
  for (int i = 0; i < 10; i++)
    if (nset[i]) aa &= ~(1 << i);
  printf("3\n^ %d\n| %d\n& %d\n", xx, oo, aa);
}
