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>
const int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-6;
const int mod = 1000000009;
using namespace std;
int n, v[500050], bit[10];
char op[500050];
int main() {
  for (int i = (int)(0); i < (int)(10); ++i) bit[i] = 2;
  scanf("%d ", &n);
  for (int i = (int)(0); i < (int)(n); ++i) {
    scanf(" %c %d", &op[i], &v[i]);
    if (op[i] == '&') {
      for (int j = (int)(0); j < (int)(10); ++j)
        if (!(v[i] & (1 << j))) bit[j] = 0;
    } else if (op[i] == '|') {
      for (int j = (int)(0); j < (int)(10); ++j)
        if (v[i] & (1 << j)) bit[j] = 1;
    } else {
      for (int j = (int)(0); j < (int)(10); ++j)
        if (v[i] & (1 << j)) {
          if (abs(bit[j]) == 2)
            bit[j] *= -1;
          else
            bit[j] = 1 - bit[j];
        }
    }
  }
  vector<pair<char, int> > res;
  int val = 0;
  for (int i = (int)(0); i < (int)(10); ++i)
    if (bit[i] != 0) val |= (1 << i);
  res.push_back({'&', val});
  val = 0;
  for (int i = (int)(0); i < (int)(10); ++i)
    if (bit[i] == 1) val |= (1 << i);
  res.push_back({'|', val});
  val = 0;
  for (int i = (int)(0); i < (int)(10); ++i)
    if (bit[i] == -2) val |= (1 << i);
  res.push_back({'^', val});
  val = 0;
  cout << res.size() << endl;
  for (int i = (int)(0); i < (int)(res.size()); ++i)
    cout << res[i].first << " " << res[i].second << endl;
  return 0;
}