#include <bits/stdc++.h>
using namespace std;
const int maxn = 5 * 1e5 + 10;
char a[maxn];
int b[maxn];
int rec[4][11];
int en1[11], en0[11] = {0};
void solve(char p, int k) {
  int bit[11] = {0};
  for (int i = 1; i <= 10; i++) {
    bit[i] = k >> (10 - i);
    k -= bit[i] * (1 << (10 - i));
  }
  if (p == '|') {
    for (int i = 1; i <= 10; i++) {
      en1[i] = en1[i] | bit[i];
    }
    for (int i = 1; i <= 10; i++) {
      en0[i] = en0[i] | bit[i];
    }
  } else if (p == '&') {
    for (int i = 1; i <= 10; i++) {
      en1[i] = en1[i] & bit[i];
    }
    for (int i = 1; i <= 10; i++) {
      en0[i] = en0[i] & bit[i];
    }
  } else {
    for (int i = 1; i <= 10; i++) {
      en1[i] = en1[i] ^ bit[i];
    }
    for (int i = 1; i <= 10; i++) {
      en0[i] = en0[i] ^ bit[i];
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  memset(rec, 0, sizeof(rec));
  for (int i = 1; i <= 10; i++) en1[i] = 1;
  for (int i = 1; i <= 10; i++) rec[2][i] = 1;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    solve(a[i], b[i]);
  }
  for (int i = 1; i <= 10; i++) {
    if (en0[i] && en1[i]) {
      rec[1][i] = 1;
    } else if (en1[i] && !en0[i]) {
      rec[1][i] = 0;
    } else if (!en1[i] && en0[i]) {
      rec[3][i] = 1;
    } else if (!en1[i] && !en0[i]) {
      rec[2][i] = 0;
    }
  }
  int A = 0, B = 0, C = 0;
  for (int i = 1; i <= 10; i++) {
    A = A + rec[1][i] * (1 << (10 - i));
  }
  for (int i = 1; i <= 10; i++) {
    B = B + rec[2][i] * (1 << (10 - i));
  }
  for (int i = 1; i <= 10; i++) {
    C = C + rec[3][i] * (1 << (10 - i));
  }
  cout << "3" << endl;
  cout << "| " << A << endl;
  cout << "& " << B << endl;
  cout << "^ " << C << endl;
}

### Prompt

In Cpp, your task is to solve 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 maxn = 5 * 1e5 + 10;
char a[maxn];
int b[maxn];
int rec[4][11];
int en1[11], en0[11] = {0};
void solve(char p, int k) {
  int bit[11] = {0};
  for (int i = 1; i <= 10; i++) {
    bit[i] = k >> (10 - i);
    k -= bit[i] * (1 << (10 - i));
  }
  if (p == '|') {
    for (int i = 1; i <= 10; i++) {
      en1[i] = en1[i] | bit[i];
    }
    for (int i = 1; i <= 10; i++) {
      en0[i] = en0[i] | bit[i];
    }
  } else if (p == '&') {
    for (int i = 1; i <= 10; i++) {
      en1[i] = en1[i] & bit[i];
    }
    for (int i = 1; i <= 10; i++) {
      en0[i] = en0[i] & bit[i];
    }
  } else {
    for (int i = 1; i <= 10; i++) {
      en1[i] = en1[i] ^ bit[i];
    }
    for (int i = 1; i <= 10; i++) {
      en0[i] = en0[i] ^ bit[i];
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  memset(rec, 0, sizeof(rec));
  for (int i = 1; i <= 10; i++) en1[i] = 1;
  for (int i = 1; i <= 10; i++) rec[2][i] = 1;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    solve(a[i], b[i]);
  }
  for (int i = 1; i <= 10; i++) {
    if (en0[i] && en1[i]) {
      rec[1][i] = 1;
    } else if (en1[i] && !en0[i]) {
      rec[1][i] = 0;
    } else if (!en1[i] && en0[i]) {
      rec[3][i] = 1;
    } else if (!en1[i] && !en0[i]) {
      rec[2][i] = 0;
    }
  }
  int A = 0, B = 0, C = 0;
  for (int i = 1; i <= 10; i++) {
    A = A + rec[1][i] * (1 << (10 - i));
  }
  for (int i = 1; i <= 10; i++) {
    B = B + rec[2][i] * (1 << (10 - i));
  }
  for (int i = 1; i <= 10; i++) {
    C = C + rec[3][i] * (1 << (10 - i));
  }
  cout << "3" << endl;
  cout << "| " << A << endl;
  cout << "& " << B << endl;
  cout << "^ " << C << endl;
}

```