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 INF = 2e9 + 5;
vector<char> get_bits(int x) {
  vector<char> ans(10, '0');
  for (int i = 0, z = 1 << 9; i < 10; i++, z >>= 1)
    if (x & z) ans[i] = '1';
  return move(ans);
}
char inverse(char c) {
  if (c == '0') return '1';
  if (c == '1') return '0';
  if (c == 'x') return 'y';
  if (c == 'y') return 'x';
}
void run_op(char op, vector<char> &status, vector<char> &x) {
  if (op == '|')
    for (int i = 0; i < 10; i++)
      if (x[i] == '1') status[i] = '1';
  if (op == '&')
    for (int i = 0; i < 10; i++)
      if (x[i] == '0') status[i] = '0';
  if (op == '^')
    for (int i = 0; i < 10; i++)
      if (x[i] == '1') status[i] = inverse(status[i]);
}
int get_char(vector<char> &x, char c) {
  int ans = 0;
  for (int i = 0, z = 512; i < 10; i++, z >>= 1)
    if (x[i] == c) ans += z;
  return ans;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> a, x, o;
  vector<char> status(10, 'x');
  while (n--) {
    char c;
    int v;
    cin >> c >> v;
    vector<char> x = get_bits(v);
    run_op(c, status, x);
  }
  int XOR = get_char(status, 'y');
  int AND = 1023 ^ get_char(status, '0');
  int OR = get_char(status, '1');
  cout << 3 << '\n';
  cout << "^ " << XOR << '\n';
  cout << "& " << AND << '\n';
  cout << "| " << OR << '\n';
}