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 + 69;
const int INF = 1e9;
int n, ans[11], mark[11];
void solve(char type, int val) {
  memset(mark, 0, sizeof mark);
  while (val != 0) {
    int lg = log2(val);
    mark[lg] = 1;
    val -= (1 << lg);
  }
  if (type == '^')
    for (int i = 0; i < 10; i++)
      if (mark[i])
        if (ans[i] == 0 or ans[i] == 2)
          ans[i]++;
        else
          ans[i]--;
  if (type == '|')
    for (int i = 0; i < 10; i++)
      if (mark[i]) ans[i] = 1;
  if (type == '&')
    for (int i = 0; i < 10; i++)
      if (!mark[i]) ans[i] = 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  char type;
  int val, XOR = 0, OR = 0;
  cin >> n;
  for (int i = 0; i < 10; i++) ans[i] = 2;
  for (int i = 0; i < n; i++) cin >> type >> val, solve(type, val);
  for (int i = 0; i < 10; i++) {
    if (ans[i] == 0) OR += (1 << i), XOR += (1 << i);
    if (ans[i] == 1) OR += (1 << i);
    if (ans[i] == 3) XOR += (1 << i);
  }
  cout << 2 << "\n";
  cout << "| " << OR << "\n";
  cout << "^ " << XOR << "\n";
}