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 long double pi = 2 * acos(0.0);
const long long prime = 31;
const long long inf = 1e18;
const long long mod = 1e18;
const long long maxn = 5e5 + 11;
long long n, d[2][11], a[maxn], l = 0, r = 1023, cnt;
char o[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> o[i] >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    if (o[i] == '^') {
      l ^= a[i];
      r ^= a[i];
    } else if (o[i] == '&') {
      l &= a[i];
      r &= a[i];
    } else {
      l |= a[i];
      r |= a[i];
    }
  }
  for (int i = 0; i < 10; ++i) {
    d[0][i] = ((l >> i) & 1);
  }
  for (int i = 0; i < 10; ++i) {
    d[1][i] = (((r >> i) & 1) ^ 1);
  }
  for (int i = 0; i < 10; ++i) {
    if (d[0][i] && d[1][i]) {
      d[0][i] = 0;
      d[1][i] = 0;
      cnt += (1 << i);
    }
  }
  cout << 3 << endl;
  cout << "^ " << cnt << endl;
  cnt = 0;
  for (int i = 0; i < 10; ++i) {
    if (d[0][i]) {
      d[0][i] = 0;
      cnt += (1 << i);
    }
  }
  cout << "| " << cnt << endl;
  cnt = 0;
  for (int i = 0; i < 10; ++i) {
    if (!d[1][i]) {
      d[1][i] = 0;
      cnt += (1 << i);
    }
  }
  cout << "& " << cnt << endl;
  cnt = 0;
  return 0;
}