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;
enum BitAbstrait { NORMAL, INVERSE, ZERO, UN };
static int maxP2[1025];
struct NombreAbstrait {
  BitAbstrait bits[10] = {NORMAL};
  void inverse(int i) {
    auto &b = bits[i];
    if (b == NORMAL)
      b = INVERSE;
    else if (b == INVERSE)
      b = NORMAL;
    else if (b == ZERO)
      b = UN;
    else if (b == UN)
      b = ZERO;
  }
  void aor(int n) {
    while (n != 0) {
      int mp = maxP2[n];
      n -= (1 << mp);
      bits[mp] = UN;
    }
  }
  void aand(int n) {
    bool bn[10] = {0};
    while (n != 0) {
      int mp = maxP2[n];
      n -= (1 << mp);
      bn[mp] = 1;
    }
    for (int i = 0; i < 10; ++i)
      if (bn[i] == 0) bits[i] = ZERO;
  }
  void axor(int n) {
    while (n != 0) {
      int mp = maxP2[n];
      n -= (1 << mp);
      inverse(mp);
    }
  }
  void dbg() {
    for (int i = 0; i < 10; ++i) {
      string s;
      if (bits[i] == 0)
        s = "NORMAL";
      else if (bits[i] == 1)
        s = "INVERSE";
      else if (bits[i] == 2)
        s = "ZERO";
      else
        s = "UN";
      cout << "bit " << i << " : " << s << "\n";
    }
  }
};
void cpre() {
  maxP2[0] = -1;
  maxP2[1] = 0;
  int np2 = 2;
  for (int i = 2; i < 1025; ++i) {
    maxP2[i] = maxP2[i - 1];
    if (i == np2) {
      ++maxP2[i], np2 *= 2;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cpre();
  NombreAbstrait nb;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    char c;
    int x;
    cin >> c >> x;
    if (c == '|') nb.aor(x);
    if (c == '&') nb.aand(x);
    if (c == '^') nb.axor(x);
  }
  int nor = 0, nxor = 0, nand = 0;
  for (int i = 0; i < 10; ++i) {
    auto &b = nb.bits[i];
    if (b == 1) nxor += (1 << i);
    if (b == 3) nor += (1 << i);
    if (b != 2) nand += (1 << i);
  }
  cout << "3\n| " << nor << "\n^ " << nxor << "\n& " << nand;
  return 0;
}