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 int N = 1e6 + 5;
int n;
pair<char, int> a[N];
int get_bit(int i, int mask) {
  if ((1 << (i)) & mask)
    return 1;
  else
    return 0;
}
void enter() {
  cin >> n;
  for (int i = (1), _b = (n); i <= _b; ++i) cin >> a[i].first >> a[i].second;
}
int change[12][2];
void process() {
  for (int i = (0), _b = (10); i < _b; ++i) {
    for (int t = (0), _b = (1); t <= _b; ++t) {
      bool cur = t;
      for (int j = (1), _b = (n); j <= _b; ++j) {
        if (a[j].first == '|') {
          cur |= get_bit(i, a[j].second);
        } else if (a[j].first == '^') {
          cur ^= get_bit(i, a[j].second);
        } else if (a[j].first == '&') {
          cur &= get_bit(i, a[j].second);
        }
      }
      change[i][t] = cur;
    }
  }
  for (int x = (0), _b = (1023); x <= _b; ++x) {
    bool ok = 1;
    int y = 0;
    for (int num = (0), _b = (1023); num <= _b; ++num) {
      int c = (num | x);
      for (int i = (0), _b = (10); i < _b; ++i) {
        if (get_bit(i, c) == 1 && change[i][get_bit(i, num)] == 0) {
          y |= (1 << (i));
        }
      }
    }
    for (int num = (0), _b = (1023); num <= _b; ++num) {
      int c = (num | x) ^ y;
      for (int i = (0), _b = (10); i < _b; ++i) {
        if (get_bit(i, c) != change[i][get_bit(i, num)]) {
          ok = 0;
        }
      }
    }
    if (ok) {
      cout << 2 << '\n';
      cout << "| " << x << '\n';
      cout << "^ " << y << '\n';
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  enter();
  process();
  return 0;
}