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 MAX_N = 1e6 + 10;
const int MAX_LOG = 20;
const int INF = 1e9 + 1;
const long long MOD = 1e9 + 7;
int n, x, o, a;
pair<char, int> q[MAX_N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> q[i].first >> q[i].second;
  }
  x = 0;
  a = 1023;
  o = 0;
  for (int i = 0; i < 10; i++) {
    bool zero = 0, one = 1;
    for (int j = 1; j <= n; j++) {
      char ch = q[j].first;
      int num = q[j].second;
      bool is = false;
      if ((1 << i) & num) is = true;
      if (ch == '^') {
        zero = zero ^ is;
        one = one ^ is;
      } else if (ch == '&') {
        zero = zero & is;
        one = one & is;
      } else if (ch == '|') {
        zero = zero | is;
        one = one | is;
      }
    }
    if (zero == 0 && one == 0) {
      a -= (1 << i);
    }
    if (zero == 1 and one == 1) {
      o += (1 << i);
    }
    if (zero == 1 and one == 0) {
      x += (1 << i);
    }
  }
  cout << 3 << endl;
  cout << "^ " << x << endl;
  cout << "& " << a << endl;
  cout << "| " << o << endl;
}