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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  vector<pair<char, int> > ans;
  vector<int> B;
  vector<char> A;
  cin >> n;
  int a1 = 0;
  int a2 = 1023;
  for (int i = 0; i < n; ++i) {
    char a;
    int b;
    cin >> a >> b;
    if (a == '|') {
      a1 = a1 | b;
      a2 = a2 | b;
    }
    if (a == '^') {
      a1 = a1 ^ b;
      a2 = a2 ^ b;
    }
    if (a == '&') {
      a1 = a1 & b;
      a2 = a2 & b;
    }
  }
  int res1 = 0;
  int res2 = 0;
  for (int i = 0; i < 10; ++i) {
    if (((1 << i) & (a2)) == 0) {
      res1 = res1 + (1 << i);
    }
  }
  if (res1 != 0) {
    ans.push_back(make_pair('^', res1));
    a1 = a1 ^ res1;
  }
  for (int i = 0; i < 10; ++i) {
    if (((1 << i) & a1) != 0) {
      res2 = res2 + (1 << i);
    }
  }
  if (res2 != 0) ans.push_back(make_pair('|', res2));
  reverse(ans.begin(), ans.end());
  cout << (int)ans.size() << endl;
  for (int i = 0; i < (int)ans.size(); ++i) {
    cout << ans[i].first << ' ' << ans[i].second << endl;
  }
  return 0;
}
