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;
char res[15][2];
map<char, vector<int> > kek;
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = 0; i < 14; i++) {
    res[i][1] = 1;
  }
  int n;
  cin >> n;
  while (n--) {
    char c;
    int a;
    cin >> c >> a;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 2; j++) {
        if (c == '^') {
          res[i][j] ^= (bool)(a & (1 << i));
        }
        if (c == '&') {
          res[i][j] &= (bool)(a & (1 << i));
        }
        if (c == '|') {
          res[i][j] |= (bool)(a & (1 << i));
        }
      }
    }
  }
  for (int i = 0; i < 10; i++) {
    if (!res[i][0] && res[i][1]) {
      continue;
    }
    if (!res[i][0] && !res[i][1]) {
      kek['&'].push_back(i);
    }
    if (res[i][0] && res[i][1]) {
      kek['|'].push_back(i);
    }
    if (res[i][0] && !res[i][1]) {
      kek['^'].push_back(i);
    }
  }
  for (int j = 0; j < 2; j++) {
    for (int i = 0; i < 10; i++) {
    }
  }
  cout << 3 << endl;
  int a1nd = 0;
  for (auto x : kek['&']) {
    a1nd |= ((1 << x));
  }
  cout << '&' << ' ' << ((~a1nd) & 1023) << endl;
  int o1r = 0;
  for (auto x : kek['|']) {
    o1r |= (1 << x);
  }
  cout << '|' << ' ' << o1r << endl;
  int x1or = 0;
  for (auto x : kek['^']) {
    x1or |= (1 << x);
  }
  cout << '^' << ' ' << x1or << endl;
  return 0;
}