#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int expo(int x, int y) {
  int res = 1;
  x = x % M;
  while (y > 0) {
    if (y & 1) res = (1ll * res * x) % M;
    y = y >> 1;
    x = (1ll * x * x) % M;
  }
  return res;
}
int ncr(int n, int r) {
  int res = 1;
  if (r > n - r) r = n - r;
  for (int i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
bool prime(long long n) {
  long long i;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    int n, i, z;
    char ch;
    cin >> n;
    vector<pair<char, int>> v;
    for (i = 0; i < n; i++) {
      cin >> ch >> z;
      v.push_back({ch, z});
    }
    int a1[10] = {}, b1[10] = {}, a2 = 0, b2 = 1023;
    for (i = 0; i < n; i++) {
      if (v[i].first == '|') {
        a2 |= v[i].second;
        b2 |= v[i].second;
      } else if (v[i].first == '&') {
        a2 &= v[i].second;
        b2 &= v[i].second;
      } else {
        a2 ^= v[i].second;
        b2 ^= v[i].second;
      }
    }
    int k = 0;
    while (a2 != 0) {
      a1[k++] = (a2 % 2);
      a2 /= 2;
    }
    k = 0;
    while (b2 != 0) {
      b1[k++] = (b2 % 2);
      b2 /= 2;
    }
    int z1 = 0, z2 = 0, z3 = 0;
    for (i = 0; i < 10; i++) {
      if (a1[i] == 1 && b1[i] == 0) z1 += pow(2, i);
      if (a1[i] == 1 && b1[i] == 1) z2 += pow(2, i);
      if (!(a1[i] == 0 && b1[i] == 0)) z3 += pow(2, i);
    }
    cout << 3 << "\n";
    cout << "^ " << z1 << "\n";
    cout << "| " << z2 << "\n";
    cout << "& " << z3 << "\n";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 M = 1e9 + 7;
int expo(int x, int y) {
  int res = 1;
  x = x % M;
  while (y > 0) {
    if (y & 1) res = (1ll * res * x) % M;
    y = y >> 1;
    x = (1ll * x * x) % M;
  }
  return res;
}
int ncr(int n, int r) {
  int res = 1;
  if (r > n - r) r = n - r;
  for (int i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
bool prime(long long n) {
  long long i;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    int n, i, z;
    char ch;
    cin >> n;
    vector<pair<char, int>> v;
    for (i = 0; i < n; i++) {
      cin >> ch >> z;
      v.push_back({ch, z});
    }
    int a1[10] = {}, b1[10] = {}, a2 = 0, b2 = 1023;
    for (i = 0; i < n; i++) {
      if (v[i].first == '|') {
        a2 |= v[i].second;
        b2 |= v[i].second;
      } else if (v[i].first == '&') {
        a2 &= v[i].second;
        b2 &= v[i].second;
      } else {
        a2 ^= v[i].second;
        b2 ^= v[i].second;
      }
    }
    int k = 0;
    while (a2 != 0) {
      a1[k++] = (a2 % 2);
      a2 /= 2;
    }
    k = 0;
    while (b2 != 0) {
      b1[k++] = (b2 % 2);
      b2 /= 2;
    }
    int z1 = 0, z2 = 0, z3 = 0;
    for (i = 0; i < 10; i++) {
      if (a1[i] == 1 && b1[i] == 0) z1 += pow(2, i);
      if (a1[i] == 1 && b1[i] == 1) z2 += pow(2, i);
      if (!(a1[i] == 0 && b1[i] == 0)) z3 += pow(2, i);
    }
    cout << 3 << "\n";
    cout << "^ " << z1 << "\n";
    cout << "| " << z2 << "\n";
    cout << "& " << z3 << "\n";
  }
  return 0;
}

```