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;
int n;
int btod(vector<int> &b) {
  int t = 1;
  int ans = 0;
  for (int i = 0; i < b.size(); i++) {
    ans += b[i] * t;
    t *= 2;
  }
  return ans;
}
vector<int> dtob(int d) {
  vector<int> b(10, 0);
  for (int i = 0; i < 10; i++) {
    b[i] = d % 2;
    d = d / 2;
    if (d == 0) break;
  }
  return b;
}
int main() {
  scanf("%d", &n);
  vector<int> zero(10, 0), one(10, 0);
  vector<int> num(n, 0);
  vector<int> op(n, 0);
  string s;
  for (int i = 0; i < n; i++) {
    cin >> s;
    if (s[0] == '|')
      op[i] = 1;
    else if (s[0] == '^')
      op[i] = 2;
    cin >> s;
    num[i] = stoi(s);
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (op[i] == 0)
      ans = ans & num[i];
    else if (op[i] == 1)
      ans = ans | num[i];
    else
      ans = ans ^ num[i];
  }
  zero = dtob(ans);
  ans = 1023;
  for (int i = 0; i < n; i++) {
    if (op[i] == 0)
      ans = ans & num[i];
    else if (op[i] == 1)
      ans = ans | num[i];
    else
      ans = ans ^ num[i];
  }
  one = dtob(ans);
  vector<int> bits(10, 0), op2(10, 0);
  vector<int> AND(10, 0), OR(10, 0), XOR(10, 0);
  for (int i = 0; i < 10; i++) {
    if (zero[i] == 0 && one[i] == 0) {
    } else if (zero[i] == 1 && one[i] == 1) {
      OR[i] = 1;
      AND[i] = 1;
      XOR[i] = 0;
    } else if (zero[i] == 1 && one[i] == 0) {
      AND[i] = 1;
      XOR[i] = 1;
    } else {
      AND[i] = 1;
    }
  }
  cout << 3 << endl;
  cout << "| " << btod(OR) << endl;
  cout << "& " << btod(AND) << endl;
  cout << "^ " << btod(XOR) << endl;
  return 0;
}