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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-7;
int d[11];
int goto2(int x) {
  for (int i = 0; i < (10); i++) d[i] = 0;
  int i = 0;
  while (x > 0) {
    d[i] = x % 2;
    x /= 2;
    i++;
  }
  return i;
}
int main() {
  int x[11];
  int n, in;
  char q;
  for (int i = 0; i < (10); i++) {
    x[i] = 4;
  }
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> q >> in;
    int kkk = goto2(in);
    if (q == '|') {
      for (int j = 0; j < 10; j++) {
        if (d[j] == 1) x[j] = 1;
      }
    }
    if (q == '^') {
      for (int j = 0; j < 10; j++) {
        if (d[j] == 1) {
          if (x[j] == 0) {
            x[j] = 1;
          } else if (x[j] == 1) {
            x[j] = 0;
          }
          if (x[j] == 4) {
            x[j] = -4;
          } else if (x[j] == -4) {
            x[j] = 4;
          }
        }
      }
    }
    if (q == '&') {
      for (int j = 0; j < 10; j++) {
        if (d[j] == 0) x[j] = 0;
      }
    }
  }
  vector<int> v1;
  vector<int> v2;
  vector<int> v3;
  for (int i = 0; i < 10; i++) {
    if (x[i] == 0) v1.push_back(i);
    if (x[i] == 1) v2.push_back(i);
    if (x[i] == -4) v3.push_back(i);
  }
  int an1d = 1023;
  if (v1.size() == 0)
    an1d = 1023;
  else {
    for (auto t : v1) {
      an1d -= (1 << t);
    }
  }
  int o1r = 0;
  if (v2.size() == 0)
    o1r = 0;
  else {
    for (auto t : v2) {
      o1r += (1 << t);
    }
  }
  int x1or = 0;
  if (v3.size() == 0)
    x1or = 0;
  else {
    for (auto t : v3) {
      x1or += (1 << t);
    }
  }
  cout << 3 << endl;
  cout << "& " << an1d << endl;
  cout << "| " << o1r << endl;
  cout << "^ " << x1or << endl;
}