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 long long llmax = 9223372036854775807;
const long long llmin = -9223372036854775808;
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0)
    return a + b;
  else
    return gcd(b, a % b);
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
int ar0[10];
int ar1[10];
int main() {
  bool test = false;
  if (test) freopen("/Users/aj/code/HackerRank/input.txt", "r", stdin);
  int n;
  cin >> n;
  int zero = 0, one = 0, o1 = 1;
  for (int i = 0; i < 10; i++) {
    ar1[i] = 1;
    one |= o1;
    o1 <<= 1;
  }
  for (int i = 0; i < n; i++) {
    char c;
    int num;
    cin >> c >> num;
    if (c == '|') {
      zero |= num;
      one |= num;
    } else if (c == '&') {
      zero &= num;
      one &= num;
    } else {
      zero ^= num;
      one ^= num;
    }
  }
  o1 = 1;
  vector<int> mor, mand, mxor;
  for (int i = 0; i < 10; i++) {
    int f1 = one & o1;
    int f0 = zero & o1;
    if (f1 == 0 && f0 == 0) {
      mor.emplace_back(0);
      mxor.emplace_back(0);
      mand.emplace_back(0);
    } else if (f1 == 0 && f0 == 1) {
      mor.emplace_back(0);
      mxor.emplace_back(1);
      mand.emplace_back(1);
    } else if (f1 == 1 && f0 == 0) {
      mor.emplace_back(0);
      mxor.emplace_back(0);
      mand.emplace_back(1);
    } else if (f1 == 1 && f0 == 1) {
      mor.emplace_back(1);
      mxor.emplace_back(0);
      mand.emplace_back(1);
    }
    one >>= 1;
    zero >>= 1;
  }
  int orr = 0, andd = 0, xorr = 0;
  for (int i = 0; i < 10; i++) {
    orr <<= 1;
    andd <<= 1;
    xorr <<= 1;
    orr += mor[9 - i];
    andd += mand[9 - i];
    xorr += mxor[9 - i];
  }
  cout << 3 << endl;
  cout << "^"
       << " " << xorr << endl;
  cout << "|"
       << " " << orr << endl;
  cout << "&"
       << " " << andd << endl;
  return 0;
}