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 maxn = 5e5 + 1e4;
int dp[10010];
pair<int, int> a[maxn];
int main() {
  int n;
  char c;
  int x;
  scanf("%d", &n);
  int zero = 0, one = 1023;
  for (int i = 0; i < n; i++) {
    getchar();
    scanf("%c", &a[i].first);
    scanf("%d", &a[i].second);
    if (a[i].first == '|') one |= a[i].second, zero |= a[i].second;
    if (a[i].first == '&') one &= a[i].second, zero &= a[i].second;
    if (a[i].first == '^') one ^= a[i].second, zero ^= a[i].second;
  }
  for (int i = 0; i < 1023; i++) {
    for (int j = 0; j < 1023; j++) {
      if (((0 & i) & j) == zero && ((1023 & i) & j) == one) {
        cout << "2\n& " << i << "\n& " << j;
        return 0;
      }
      if (((0 | i) | j) == zero && ((1023 | i) | j) == one) {
        cout << "2\n| " << i << "\n| " << j;
        return 0;
      }
      if (((0 ^ i) ^ j) == zero && ((1023 ^ i) ^ j) == one) {
        cout << "2\n^ " << i << "\n^ " << j;
        return 0;
      }
      if (((0 & i) | j) == zero && ((1023 & i) | j) == one) {
        cout << "2\n& " << i << "\n| " << j;
        return 0;
      }
      if (((0 & i) ^ j) == zero && ((1023 & i) ^ j) == one) {
        cout << "2\n& " << i << "\n^ " << j;
        return 0;
      }
      if (((0 | i) ^ j) == zero && ((1023 | i) ^ j) == one) {
        cout << "2\n| " << i << "\n^ " << j;
        return 0;
      }
      if (((0 | i) & j) == zero && ((1023 | i) & j) == one) {
        cout << "2\n| " << i << "\n& " << j;
        return 0;
      }
      if (((0 ^ i) | j) == zero && ((1023 ^ i) | j) == one) {
        cout << "2\n^ " << i << "\n| " << j;
        return 0;
      }
      if (((0 ^ i) & j) == zero && ((1023 ^ i) & j) == one) {
        cout << "2\n^ " << i << "\n& " << j;
        return 0;
      }
    }
  }
  cout << 0;
  return 0;
}