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 = 0x3f3f3f3f;
template <class T>
inline int len(const T &a) {
  return a.size();
}
const double EPS = 1e-9;
const double PI = 3.14159265358979323846264338328;
const long long LINF = (long long)1e16 + 5;
template <class T>
inline int toInt(const T &t) {
  int n;
  stringstream ss;
  ss << t;
  ss >> n;
  return n;
}
template <class T>
inline string toStr(const T &t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
template <class T>
inline long long toInt64(const T &t) {
  long long n;
  stringstream ss;
  ss << t;
  ss >> n;
  return n;
}
template <typename T>
int sgn(T val) {
  return (T(0) < val) - (val < T(0));
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
  bool first = true;
  os << "[";
  for (unsigned int i = 0; i < v.size(); i++) {
    if (!first) os << ", ";
    os << v[i];
    first = false;
  }
  return os << "]";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::set<T> &S) {
  bool first = true;
  os << "set([";
  for (typename std::set<T>::const_iterator it = S.begin(); it != S.end();
       ++it) {
    if (!first) os << ", ";
    os << *it;
    first = false;
  }
  return os << "])";
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::map<T1, T2> &d) {
  bool first = true;
  os << "{";
  for (typename std::map<T1, T2>::const_iterator it = d.begin(); it != d.end();
       ++it) {
    if (!first) os << ", ";
    os << it->first << ": " << it->second;
    first = false;
  }
  return os << "}";
}
struct debugger {
  template <typename T>
  debugger &operator,(const T v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
int main() {
  int N, n;
  char buf[55];
  int l[10] = {2, 2, 2, 2, 2, 2, 2, 2, 2, 2};
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%s %d", buf, &n);
    if (buf[0] == '|') {
      for (int j = 0; j < 10; j++)
        if ((n >> j) & 1) l[10 - j - 1] = 1;
    } else if (buf[0] == '&') {
      for (int j = 0; j < 10; j++)
        if (!((n >> j) & 1)) l[10 - j - 1] = 0;
    } else if (buf[0] == '^') {
      for (int j = 0; j < 10; j++)
        if ((n >> j) & 1) l[10 - j - 1] ^= 1;
    }
  }
  int a = 1023, b = 0, c = 0;
  for (int i = 0; i < 10; i++) {
    int x = 10 - i - 1;
    if (l[i] == 2) continue;
    if (l[i] == 0) a &= ~(1 << x);
    if (l[i] == 1) b |= 1 << x;
    if (l[i] == 3) c ^= 1 << x;
  }
  int ret = 0;
  if (a != 1023) ret++;
  if (b != 0) ret++;
  if (c != 0) ret++;
  printf("%d\n", ret);
  if (a != 1023) printf("& %d\n", a);
  if (b != 0) printf("| %d\n", b);
  if (c != 0) printf("^ %d\n", c);
  return 0;
}