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 = 5e5 + 10, MAXN = 1e4 + 10, MOD = 1e9 + 7;
char c[N];
int a[N];
vector<pair<long long, long long> > v, v1;
int main() {
  ios::sync_with_stdio(0);
  int n;
  int ret1 = 1023, ret2 = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> c[i] >> a[i];
    if (c[i] == '|') {
      ret1 |= a[i];
      ret2 |= a[i];
    } else if (c[i] == '&') {
      ret1 &= a[i];
      ret2 &= a[i];
    } else {
      ret1 ^= a[i];
      ret2 ^= a[i];
    }
  }
  int ans1 = 0, ans2 = 1023, ans3 = 0, ans4 = 0;
  for (int i = 0; i < 10; i++) {
    if (((1 << i) & ret1) == ((1 << i) & ret2) && ((1 << i) & ret1)) {
      ans1 |= (1 << i);
    } else if (((1 << i) & ret1) == ((1 << i) & ret2) && !((1 << i) & ret1)) {
      if (ans2 & (1 << i)) ans2 ^= (1 << i);
    } else if (((1 << i) & ret1)) {
    } else {
      ans4 ^= (1 << i);
    }
  }
  cout << 3 << endl;
  cout << "| " << ans1 << endl << "& " << ans2 << endl << "^ " << ans4;
}