#include <bits/stdc++.h>
using namespace std;
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
const long double pi = acos(-1);
char C[500500];
int X[500500];
int op_and[20];
int op_xor[20];
int op_or[20];
int apply(int bit, int pos, char op, int x) {
  if (op == '&') {
    if (x & (1 << pos))
      return bit & 1;
    else
      return 0;
  } else if (op == '|') {
    if (x & (1 << pos))
      return 1;
    else
      return bit;
  } else if (op == '^') {
    if (x & (1 << pos))
      return bit ^ 1;
    else
      return bit;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout << fixed;
  cout.tie(0);
  cin.tie(0);
  int N;
  cin >> N;
  for (int(i) = 0; (i) < (N); (i)++) cin >> C[i] >> X[i];
  for (int(i) = 0; (i) < (20); (i)++) op_and[i] = -1;
  for (int(i) = 0; (i) < (20); (i)++) op_xor[i] = -1;
  for (int(i) = 0; (i) < (20); (i)++) op_or[i] = -1;
  for (int(j) = 0; (j) < (10); (j)++) {
    int bit0 = 0;
    for (int(i) = 0; (i) < (N); (i)++) {
      bit0 = apply(bit0, j, C[i], X[i]);
    }
    int bit1 = 1;
    for (int(i) = 0; (i) < (N); (i)++) {
      bit1 = apply(bit1, j, C[i], X[i]);
    }
    if (bit0 == 0 && bit1 == 0) op_and[j] = 0;
    if (bit0 == 0 && bit1 == 1) op_xor[j] = 0;
    if (bit0 == 1 && bit1 == 0) op_xor[j] = 1;
    if (bit0 == 1 && bit1 == 1) op_or[j] = 1;
  }
  int andnum = 0;
  int xornum = 0;
  int ornum = 0;
  for (int(j) = 0; (j) < (10); (j)++) {
    if (op_and[j] != 0) andnum += 1 << j;
    if (op_xor[j] == 1) xornum += 1 << j;
    if (op_or[j] == 1) ornum += 1 << j;
  }
  cout << 3 << endl;
  cout << "& " << andnum << endl;
  cout << "| " << ornum << endl;
  cout << "^ " << xornum << endl;
}

### Prompt

Create a solution in cpp for 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 = INT_MAX;
const long long INFL = LLONG_MAX;
const long double pi = acos(-1);
char C[500500];
int X[500500];
int op_and[20];
int op_xor[20];
int op_or[20];
int apply(int bit, int pos, char op, int x) {
  if (op == '&') {
    if (x & (1 << pos))
      return bit & 1;
    else
      return 0;
  } else if (op == '|') {
    if (x & (1 << pos))
      return 1;
    else
      return bit;
  } else if (op == '^') {
    if (x & (1 << pos))
      return bit ^ 1;
    else
      return bit;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout << fixed;
  cout.tie(0);
  cin.tie(0);
  int N;
  cin >> N;
  for (int(i) = 0; (i) < (N); (i)++) cin >> C[i] >> X[i];
  for (int(i) = 0; (i) < (20); (i)++) op_and[i] = -1;
  for (int(i) = 0; (i) < (20); (i)++) op_xor[i] = -1;
  for (int(i) = 0; (i) < (20); (i)++) op_or[i] = -1;
  for (int(j) = 0; (j) < (10); (j)++) {
    int bit0 = 0;
    for (int(i) = 0; (i) < (N); (i)++) {
      bit0 = apply(bit0, j, C[i], X[i]);
    }
    int bit1 = 1;
    for (int(i) = 0; (i) < (N); (i)++) {
      bit1 = apply(bit1, j, C[i], X[i]);
    }
    if (bit0 == 0 && bit1 == 0) op_and[j] = 0;
    if (bit0 == 0 && bit1 == 1) op_xor[j] = 0;
    if (bit0 == 1 && bit1 == 0) op_xor[j] = 1;
    if (bit0 == 1 && bit1 == 1) op_or[j] = 1;
  }
  int andnum = 0;
  int xornum = 0;
  int ornum = 0;
  for (int(j) = 0; (j) < (10); (j)++) {
    if (op_and[j] != 0) andnum += 1 << j;
    if (op_xor[j] == 1) xornum += 1 << j;
    if (op_or[j] == 1) ornum += 1 << j;
  }
  cout << 3 << endl;
  cout << "& " << andnum << endl;
  cout << "| " << ornum << endl;
  cout << "^ " << xornum << endl;
}

```