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;
template <typename T, typename U>
bool ckmin(T &a, U b) {
  return b < a ? a = b, true : false;
}
template <typename T, typename U>
bool ckmax(T &a, U b) {
  return b > a ? a = b, true : false;
}
int n;
map<char, int> m;
vector<vector<int>> ans;
vector<pair<char, int>> op;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    char x;
    int y;
    cin >> x >> y;
    op.push_back({x, y});
  }
  m['&'] = 1023;
  for (int i = 0; i < 10; ++i) {
    int cur = (1 << i), res = 0;
    for (int j = 0; j < n; ++j) {
      if (op[j].first == '|') cur |= op[j].second;
      if (op[j].first == '&') cur &= op[j].second;
      if (op[j].first == '^') cur ^= op[j].second;
    }
    res |= (cur >> i & 1);
    res <<= 1;
    cur = 0;
    for (int j = 0; j < n; ++j) {
      if (op[j].first == '|') cur |= op[j].second;
      if (op[j].first == '&') cur &= op[j].second;
      if (op[j].first == '^') cur ^= op[j].second;
    }
    res |= (cur >> i & 1);
    if (res == 3) m['|'] |= (1 << i);
    if (res == 0) m['&'] ^= (1 << i);
    if (res == 1) m['^'] |= (1 << i);
  }
  cout << (int)(m).size() << "\n";
  for (auto i : m) cout << i.first << " " << i.second << "\n";
  return 0;
}
