#include <bits/stdc++.h>
using namespace std;
int n;
int x[512345];
char c[512345];
int d[5][20];
vector<int> And;
vector<int> Or;
vector<int> Xor;
int t;
int num;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i] >> x[i];
  for (int k = 0; k <= 9; k++) {
    int ans = 1 << k;
    for (int i = 1; i <= n; i++) {
      if (c[i] == '&') ans &= x[i];
      if (c[i] == '|') ans |= x[i];
      if (c[i] == '^') ans ^= x[i];
    }
    d[1][k] = (ans >> k) & 1;
  }
  for (int k = 0; k <= 9; k++) {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (c[i] == '&') ans &= x[i];
      if (c[i] == '|') ans |= x[i];
      if (c[i] == '^') ans ^= x[i];
    }
    d[0][k] = (ans >> k) & 1;
  }
  for (int k = 0; k <= 9; k++) {
    if (d[0][k] == 1 && d[1][k] == 0)
      Xor.push_back(k);
    else if (d[0][k] == 1)
      Or.push_back(k);
    else if (d[1][k] == 0)
      And.push_back(k);
  }
  t = !Or.empty() + !Xor.empty() + !And.empty();
  cout << t << '\n';
  num = 0;
  if (!Or.empty()) {
    for (auto i : Or) num += 1 << i;
    cout << '|' << ' ' << num << '\n';
  }
  num = 0;
  if (!And.empty()) {
    for (auto i : And) num += 1 << i;
    num = 1023 - num;
    cout << '&' << ' ' << num << '\n';
  }
  num = 0;
  if (!Xor.empty()) {
    for (auto i : Xor) num += 1 << i;
    cout << '^' << ' ' << num << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int n;
int x[512345];
char c[512345];
int d[5][20];
vector<int> And;
vector<int> Or;
vector<int> Xor;
int t;
int num;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i] >> x[i];
  for (int k = 0; k <= 9; k++) {
    int ans = 1 << k;
    for (int i = 1; i <= n; i++) {
      if (c[i] == '&') ans &= x[i];
      if (c[i] == '|') ans |= x[i];
      if (c[i] == '^') ans ^= x[i];
    }
    d[1][k] = (ans >> k) & 1;
  }
  for (int k = 0; k <= 9; k++) {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      if (c[i] == '&') ans &= x[i];
      if (c[i] == '|') ans |= x[i];
      if (c[i] == '^') ans ^= x[i];
    }
    d[0][k] = (ans >> k) & 1;
  }
  for (int k = 0; k <= 9; k++) {
    if (d[0][k] == 1 && d[1][k] == 0)
      Xor.push_back(k);
    else if (d[0][k] == 1)
      Or.push_back(k);
    else if (d[1][k] == 0)
      And.push_back(k);
  }
  t = !Or.empty() + !Xor.empty() + !And.empty();
  cout << t << '\n';
  num = 0;
  if (!Or.empty()) {
    for (auto i : Or) num += 1 << i;
    cout << '|' << ' ' << num << '\n';
  }
  num = 0;
  if (!And.empty()) {
    for (auto i : And) num += 1 << i;
    num = 1023 - num;
    cout << '&' << ' ' << num << '\n';
  }
  num = 0;
  if (!Xor.empty()) {
    for (auto i : Xor) num += 1 << i;
    cout << '^' << ' ' << num << '\n';
  }
  return 0;
}

```