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() {
  int n;
  int maxbit = -1;
  cin >> n;
  char c[n];
  int bit[n][10];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    int x;
    cin >> x;
    for (int j = 0; j < 10; j++) {
      bit[i][j] = (x >> j) & 1;
      if (bit[i][j] == 1 && j > maxbit) maxbit = j;
    }
  }
  int fin[10][2];
  for (int i = 0; i <= 9; i++) {
    fin[i][0] = 0;
    fin[i][1] = 1;
    for (int j = 0; j < n; j++) {
      if (c[j] == '|') {
        fin[i][0] |= bit[j][i];
        fin[i][1] |= bit[j][i];
      } else if (c[j] == '&') {
        fin[i][0] &= bit[j][i];
        fin[i][1] &= bit[j][i];
      } else if (c[j] == '^') {
        fin[i][0] ^= bit[j][i];
        fin[i][1] ^= bit[j][i];
      }
    }
  }
  int orr[10], andd[10], xorr[10];
  for (int i = 0; i <= 9; i++) {
    orr[i] = 0;
    andd[i] = 1;
    xorr[i] = 0;
  }
  for (int i = 0; i < 10; i++) {
    if (fin[i][0] == 0 && fin[i][1] == 0) {
      andd[i] = 0;
    } else if (fin[i][0] == 1 && fin[i][1] == 0) {
      xorr[i] = 1;
    } else if (fin[i][0] == 1 && fin[i][1] == 1) {
      orr[i] = 1;
    }
  }
  int o = 0, xo = 0, an = 0;
  for (int i = 0; i < 10; i++) {
    o += orr[i] * pow(2, i);
    xo += xorr[i] * pow(2, i);
    an += andd[i] * pow(2, i);
  }
  int x = 0, y = 0, z = 0;
  if (o > 0) x = 1;
  if (an < 1023) y = 1;
  if (xo > 0) z = 1;
  int cnt = x + y + z;
  cout << cnt << endl;
  if (o > 0) cout << '|' << " " << o << endl;
  if (an < 1023) cout << "& " << an << endl;
  if (xo > 0) cout << "^ " << xo << endl;
  return 0;
}