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.
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5;
int c0[11], c1[11], a[3][N], n, xo, an = 1023, o;
int f1(int k) {
  int el = 1 << k;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 3; ++j) {
      if (a[j][i] != -1) {
        if (!j)
          el |= a[j][i];
        else if (j == 1)
          el &= a[j][i];
        else
          el ^= a[j][i];
      }
    }
  }
  return (el & (1 << k)) != 0;
}
int f0(int k) {
  int el = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 3; ++j) {
      if (a[j][i] != -1) {
        if (!j)
          el |= a[j][i];
        else if (j == 1)
          el &= a[j][i];
        else
          el ^= a[j][i];
      }
    }
  }
  return (el & (1 << k)) != 0;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    char ch;
    int c;
    cin >> ch >> c;
    a[0][i] = -1, a[1][i] = -1, a[2][i] = -1;
    if (ch == '|')
      a[0][i] = c;
    else if (ch == '&')
      a[1][i] = c;
    else
      a[2][i] = c;
  }
  for (int i = 0; i <= 9; ++i) {
    c1[i] = f1(i);
    c0[i] = f0(i);
    if (c1[i] && c0[i])
      o |= 1 << i;
    else if (!c1[i] && !c0[i])
      an -= 1 << i;
    else if (!c1[i] && c0[i])
      xo |= 1 << i;
  }
  cout << "3\n";
  cout << "& " << an << '\n' << "| " << o << '\n' << "^ " << xo;
}
