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 INF = 0x3f3f3f3f;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
long long nxt() {
  long long x;
  cin >> x;
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (auto& [x, y] : a) {
    char c;
    cin >> c >> y;
    if (c == '&')
      x = 1;
    else if (c == '|')
      x = 2;
    else
      x = 3;
  }
  int an = (1 << 10) - 1, o = 0, xo = 0;
  for (int pos = 0; pos < int(10); ++pos) {
    int ans[2] = {0};
    for (int ini : {0, 1}) {
      int res = ini;
      for (auto [x, y] : a) {
        if (x == 1)
          res &= y >> pos & 1;
        else if (x == 2)
          res |= y >> pos & 1;
        else
          res ^= y >> pos & 1;
      }
      ans[ini] = res;
    }
    if (ans[0] == 0 && ans[1] == 0)
      an -= 1 << pos;
    else if (ans[0] == 1 && ans[1] == 0)
      xo += 1 << pos;
    else if (ans[0] == 1 && ans[1] == 1)
      o += 1 << pos;
  }
  cout << 3 << '\n';
  cout << "& " << an << "\n| " << o << "\n^ " << xo;
  return 0;
}