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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long n;
  string x = "??????????";
  cin >> n;
  char arr[n];
  int a[n], y;
  vector<int> v;
  string ans = x;
  for (int i = 0; i < n; ++i) {
    v.resize(10);
    v[0] = v[1] = v[2] = v[3] = v[4] = v[5] = v[6] = v[7] = v[8] = v[9] = 0;
    cin >> arr[i] >> a[i];
    int j = 0;
    y = a[i];
    while (y > 0) {
      v[9 - j] = y % 2;
      y /= 2;
      j++;
    }
    if (arr[i] == '|') {
      for (int j = 0; j < 10; ++j) {
        if (v[j] == 1) ans[j] = '1';
      }
    } else if (arr[i] == '&') {
      for (int j = 0; j < 10; ++j) {
        if (v[j] == 0) ans[j] = '0';
      }
    } else if (arr[i] == '^') {
      for (int j = 0; j < 10; ++j) {
        if (ans[j] == '?') {
          if (v[j] == 1) {
            ans[j] = '.';
          } else
            ans[j] = '?';
        } else if (ans[j] == '.') {
          if (v[j] == 1)
            ans[j] = '?';
          else
            ans[j] = '.';
        } else if (ans[j] == '0') {
          if (v[j] == 1)
            ans[j] = '1';
          else
            ans[j] = '0';
        } else if (ans[j] == '1') {
          if (v[j] == 1)
            ans[j] = '0';
          else
            ans[j] = '1';
        }
      }
    }
  }
  string ans1 = "1111111111";
  string ans2 = "0000000000";
  string ans3 = "0000000000";
  int ans11 = 0, ans22 = 0, ans33 = 0;
  for (int i = 0; i < 10; ++i) {
    if (ans[i] == '0') {
      ans1[i] = '0';
    } else if (ans[i] == '1') {
      ans2[i] = '1';
    } else if (ans[i] == '?') {
      ans3[i] = '0';
    } else
      ans3[i] = '1';
    ans11 += ((ans1[i] - '0') * pow(2, 9 - i));
    ans22 += ((ans2[i] - '0') * pow(2, 9 - i));
    ans33 += ((ans3[i] - '0') * pow(2, 9 - i));
  }
  cout << "3\n";
  cout << "& " << ans11 << "\n";
  cout << "| " << ans22 << "\n";
  cout << "^ " << ans33 << "\n";
  return 0;
}