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 expect[10][2] = {0};
char cmd[] = "|^&";
vector<char> ansc;
vector<int> ansa;
bool solve_rec(int depth, int max_depth, int p, vector<char> &c,
               vector<int> &a) {
  if (depth == max_depth) {
    {
      int m = 1 << p;
      for (int y = 0; y <= m; y += m) {
        int x = y;
        for (int i = 0; i < depth; ++i) {
          if (c[i] == '|') x = x | a[i];
          if (c[i] == '^') x = x ^ a[i];
          if (c[i] == '&') x = x & a[i];
        }
        if (expect[p][y / m] != (x & m)) return false;
      }
    }
    ansc = c;
    ansa = a;
    return true;
  }
  for (int i = 0; i < 2; ++i) {
    a[depth] |= (i << p);
    bool res = solve_rec(depth + 1, max_depth, p, c, a);
    if (res) return true;
    a[depth] &= ~(i << p);
  }
  return false;
}
bool solve(int max_depth, vector<char> &c) {
  vector<int> a(c.size(), 0);
  bool res = 1;
  for (int p = 0; p < 10; ++p) res &= solve_rec(0, max_depth, p, c, a);
  return res;
}
bool req_op(int depth, vector<char> &c) {
  if (depth) {
    bool res = solve(depth, c);
    if (res) return true;
  }
  if (5 <= depth) return false;
  for (int i = 0; i < 3; ++i) {
    c.push_back(cmd[i]);
    bool res = req_op(depth + 1, c);
    if (res) return true;
    c.pop_back();
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<char> c(n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> c[i] >> a[i];
  for (int p = 0; p < 10; ++p) {
    int m = 1 << p;
    for (int y = 0; y <= m; y += m) {
      int x = y;
      for (int i = 0; i < n; ++i) {
        if (c[i] == '|') x = x | a[i];
        if (c[i] == '^') x = x ^ a[i];
        if (c[i] == '&') x = x & a[i];
      }
      expect[p][y / m] = (x & m);
    }
  }
  vector<char> tmp;
  req_op(0, tmp);
  cout << ansc.size() << "\n";
  for (int i = 0; i < ansc.size(); ++i) {
    cout << ansc[i] << " " << ansa[i] << "\n";
  }
  return 0;
}