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 MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
void setmin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void setmax(T& a, T val) {
  if (a < val) a = val;
}
void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  pair<char, int> pp[n];
  for (int i = (0); i < (n); i++) cin >> pp[i].first >> pp[i].second;
  int num1 = 0, num2 = 1023;
  for (int i = (0); i < (n); i++) {
    if (pp[i].first == '|') {
      num1 |= pp[i].second;
      num2 |= pp[i].second;
    }
    if (pp[i].first == '&') {
      num1 &= pp[i].second;
      num2 &= pp[i].second;
    }
    if (pp[i].first == '^') {
      num1 ^= pp[i].second;
      num2 ^= pp[i].second;
    }
  }
  int val1 = num1 & num2;
  int val2 = (num1) & (~num2);
  int val3 = 0;
  for (int i = (0); i < (10); i++) {
    if ((num1 & (1 << i)) == 0 && (num2 & (1 << i)) == 0) val3 |= (1 << i);
  }
  val3 = 1023 & (~val3);
  cout << 3 << "\n";
  cout << "| " << val1 << "\n";
  cout << "^ " << val2 << "\n";
  cout << "& " << val3 << "\n";
  return 0;
}