#include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30;
const long long INF = 1ll << 60;
const double Inf = 1e20;
const double eps = 1e-9;
void gmax(int &a, int b) { a = (a > b ? a : b); }
void gmin(int &a, int b) { a = (a < b ? a : b); }
const int maxn = 500050;
char op[maxn][2];
int constant[maxn];
int n, final[10], x, y;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%s%d", op[i], &constant[i]);
  for (int i = 0; i <= 9; ++i) {
    x = 0;
    y = (1 << i);
    for (int i = 1; i <= n; ++i) {
      if (op[i][0] == '|') {
        x |= constant[i];
        y |= constant[i];
      }
      if (op[i][0] == '&') {
        x &= constant[i];
        y &= constant[i];
      }
      if (op[i][0] == '^') {
        x ^= constant[i];
        y ^= constant[i];
      }
    }
    x &= (1 << i);
    y &= (1 << i);
    if (x != 0 && y != 0)
      final[i] = 1;
    else if (x == 0 && y == 0)
      final[i] = 2;
    else if (x != 0 && y == 0)
      final[i] = 3;
    else
      final[i] = 4;
  }
  printf("3\n");
  int x = (1 << 10) - 1;
  for (int i = 0; i <= 9; ++i)
    if (final[i] == 2) x -= (1 << i);
  printf("& %d\n", x);
  x = 0;
  for (int i = 0; i <= 9; ++i)
    if (final[i] == 3) x += (1 << i);
  printf("^ %d\n", x);
  x = 0;
  for (int i = 0; i <= 9; ++i)
    if (final[i] == 1) x += (1 << i);
  printf("| %d\n", x);
  return 0;
}

### Prompt

Generate a Cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30;
const long long INF = 1ll << 60;
const double Inf = 1e20;
const double eps = 1e-9;
void gmax(int &a, int b) { a = (a > b ? a : b); }
void gmin(int &a, int b) { a = (a < b ? a : b); }
const int maxn = 500050;
char op[maxn][2];
int constant[maxn];
int n, final[10], x, y;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%s%d", op[i], &constant[i]);
  for (int i = 0; i <= 9; ++i) {
    x = 0;
    y = (1 << i);
    for (int i = 1; i <= n; ++i) {
      if (op[i][0] == '|') {
        x |= constant[i];
        y |= constant[i];
      }
      if (op[i][0] == '&') {
        x &= constant[i];
        y &= constant[i];
      }
      if (op[i][0] == '^') {
        x ^= constant[i];
        y ^= constant[i];
      }
    }
    x &= (1 << i);
    y &= (1 << i);
    if (x != 0 && y != 0)
      final[i] = 1;
    else if (x == 0 && y == 0)
      final[i] = 2;
    else if (x != 0 && y == 0)
      final[i] = 3;
    else
      final[i] = 4;
  }
  printf("3\n");
  int x = (1 << 10) - 1;
  for (int i = 0; i <= 9; ++i)
    if (final[i] == 2) x -= (1 << i);
  printf("& %d\n", x);
  x = 0;
  for (int i = 0; i <= 9; ++i)
    if (final[i] == 3) x += (1 << i);
  printf("^ %d\n", x);
  x = 0;
  for (int i = 0; i <= 9; ++i)
    if (final[i] == 1) x += (1 << i);
  printf("| %d\n", x);
  return 0;
}

```