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;
long long MOD = 1e9 + 7;
long long PRIME = 102838201;
long long INF = LLONG_MAX / 4;
vector<string> &split(const std::string &s, char delim, vector<string> &e) {
  stringstream ss(s);
  string item;
  while (getline(ss, item, delim)) e.push_back(item);
  return e;
}
long long Pow(long long a, long long b, long long Mo) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % Mo)
    if (b & 1) ans = ans * a % Mo;
  return ans;
}
long long nCr(long long n, long long r) {
  static long long MAXF = 1e6;
  static std::vector<long long> fact(MAXF, 1);
  for (int i = 1; i < MAXF; ++i) fact[i] = (fact[i - 1] * i) % MOD;
  MAXF = 0;
  return (fact[n] * Pow((fact[r] * fact[n - r]) % MOD, MOD - 2, MOD)) % MOD;
}
vector<int> Zfunc(string &s) {
  int n = s.length();
  vector<int> z(n, 0);
  for (int i = 1, l = 0, r = 0; i < n; i++) {
    if (i <= r) z[i] = min(z[i - l], r - i + 1);
    while (i + z[i] < n && s[i + z[i]] == s[z[i]]) z[i]++;
    if (r < i + z[i] - 1) l = i, r = i + z[i] - 1;
  }
  return z;
}
long long solve() { return 0; }
int main(int argc, char const *argv[]) {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  std::vector<int> op(10, 0);
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    int x;
    cin >> x;
    for (int j = 0; j < 10; j++) {
      if (x & (1 << j)) {
        if (c == '|') {
          op[j] = 2;
        } else if (c == '^') {
          op[j] ^= 1;
        }
      } else {
        if (c == '&') op[j] = 3;
      }
    }
  }
  cout << 3 << '\n';
  {
    int x = 0;
    for (int j = 0; j < 10; j++)
      if (op[j] == 1) x |= 1 << j;
    cout << "^ " << x << '\n';
  }
  {
    int x = 0;
    for (int j = 0; j < 11; j++)
      if (op[j] == 2) x |= 1 << j;
    cout << "| " << x << '\n';
  }
  {
    int x = 1023;
    for (int j = 0; j < 11; j++)
      if (op[j] == 3) x ^= 1 << j;
    cout << "& " << x << '\n';
  }
  return 0;
}
