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;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int INT_INF = (int)(2e9);
const ll LL_INF = (ll)(2e18);
const ll mod = 1e9 + 7;
const int NIL = -1;
static mt19937 _g(time(nullptr));
inline ll randint(ll a, ll b) {
  ll w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T, typename S>
inline ostream& operator<<(ostream& os, const pair<T, S> p) {
  cout << "[" << p.first << ";" << p.second << "]";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
const int N_BITS = 10;
vector<pair<char, int>> cmd[N_BITS];
int n;
inline int calc(int j, int input) {
  int output = input;
  for (auto& c : cmd[j]) {
    if (c.first == '^')
      output ^= c.second;
    else if (c.first == '|')
      output |= c.second;
    else
      output &= c.second;
  }
  return output;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    char op;
    int value;
    cin >> op >> value;
    for (int j = 0; j < N_BITS; ++j) cmd[j].emplace_back(op, (value >> j) & 1);
  }
  int zero = (1 << N_BITS) - 1, one = 0, inv = 0;
  for (int j = 0; j < N_BITS; ++j) {
    int f0 = calc(j, 0), f1 = calc(j, 1);
    if (f0 == 0 && f1 == 0)
      zero -= 1 << j;
    else if (f0 == 1 && f1 == 1)
      one += 1 << j;
    else if (f0 == 1 && f1 == 0)
      inv += 1 << j;
  }
  cout << 3 << '\n';
  cout << "& " << zero << '\n';
  cout << "| " << one << '\n';
  cout << "^ " << inv << '\n';
}
int main() {
  fast_io();
  solve();
  return 0;
}