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 INF = 2e9 + 999;
const long long LINF = 2e18 + 999;
void inv(bool e);
template <typename T>
T sign(T a);
template <typename T>
void die(T &a);
template <typename T>
void umax(T &a, T b);
template <typename T>
void umin(T &a, T b);
bool brute, alter;
int cnt_tests = 1;
int n;
void inp() { cin >> n; }
void solve() {
  int zero = 0, all1 = 1023;
  char type;
  int val;
  for (int i = 0; i < n; ++i) {
    cin >> type >> val;
    if (type == '|') {
      zero |= val;
      all1 |= val;
    }
    if (type == '&') {
      zero &= val;
      all1 &= val;
    }
    if (type == '^') {
      zero ^= val;
      all1 ^= val;
    }
  }
  int XOR = 0, OR = 0, AND = 1023;
  for (int i = 9; i >= 0; --i) {
    if (((zero >> i) & 1) && !((all1 >> i) & 1)) {
      XOR = (XOR | (1 << i));
    } else if (((zero >> i) & 1)) {
      OR = (OR | (1 << i));
    } else if (!((all1 >> i) & 1)) {
      AND = (AND ^ (1 << i));
    }
  }
  cout << 3 << '\n';
  cout << "| " << OR << '\n';
  cout << "^ " << XOR << '\n';
  cout << "& " << AND << '\n';
}
void stress() {}
void naive() {}
void run();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (0) {
    freopen("palindrome.in", "r", stdin);
    freopen("palindrome.out", "w", stdout);
  }
  srand(time(0));
  brute = false;
  for (int i = 0; (i < cnt_tests); i++) {
    run();
  }
  return 0;
}
void run() {
  if (!brute) {
    inp();
  } else {
    stress();
  }
  solve();
  cout << '\n';
}
template <typename T>
void die(T &a) {
  cout << a;
  exit(0);
}
template <typename T>
T sign(T a) {
  if (a == 0) return 0;
  if (a < 0) return -1;
  return 1;
}
template <typename T>
void umax(T &a, T b) {
  if (b > a) a = b;
}
template <typename T>
void umin(T &a, T b) {
  if (b < a) a = b;
}
void inv(bool e) {
  if (!e) {
    vector<int> a;
    a[-1] += 1;
  }
}