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 MAXN = 300008;
const long long inf = 1000000001;
const long long INF = 2e18;
const long double EPS = 1e-7;
const long long mod = inf + 6;
const int moff = 998244353;
const int mod13 = 1000003;
const long double PI = acos(-1.0);
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  vector<pair<char, int>> op(n);
  string kek;
  for (int i = 0; i < n; ++i) {
    string lol;
    cin >> lol;
    cin >> op[i].second;
    op[i].first = lol[0];
  }
  int orn = 0, xorn = 0, andn = 0;
  for (int bt = 0; bt < 10; ++bt) {
    int x = 0, y = 1;
    for (int i = 0; i < n; ++i) {
      if (op[i].first == '|' && (op[i].second & (1 << bt))) {
        x = y = 1;
      } else if (op[i].first == '^' && (op[i].second & (1 << bt))) {
        x ^= 1;
        y ^= 1;
      } else if (op[i].first == '&' && (op[i].second & (1 << bt))) {
        x &= 1, y &= 1;
      } else if (op[i].first == '&' && !(op[i].second & (1 << bt))) {
        x = 0, y = 0;
      }
    }
    if (x == 0 && y == 0) {
    } else {
      andn += (1 << bt);
      if (x == y) {
        orn += (1 << bt);
      } else if (x == 1) {
        xorn += (1 << bt);
      }
    }
  }
  cout << 3 << '\n';
  cout << '&' << ' ' << andn << '\n'
       << '|' << ' ' << orn << '\n'
       << '^' << ' ' << xorn << '\n';
}