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;
struct mystruct {
  long long v;
  mystruct *next;
};
int structcmp(const void *v1, const void *v2) {
  return ((mystruct *)v1)->v - ((mystruct *)v2)->v;
}
int intcmp(const void *v1, const void *v2) { return *(int *)v1 - *(int *)v2; }
long long minn(long long v1, long long v2) { return v1 < v2 ? v1 : v2; }
long long maxx(long long v1, long long v2) { return v1 > v2 ? v1 : v2; }
void getbit(int v0, int v1, int sft, int &t0, int &t1, int &t2) {
  int res = ((v0 >> sft) & 1) * 2 + ((v1 >> sft) & 1);
  switch (res) {
    case 0:
      t0 = 0;
      t1 = 0;
      t2 = 0;
      break;
    case 1:
      t0 = 0;
      t1 = 1;
      t2 = 0;
      break;
    case 2:
      t0 = 1;
      t1 = 1;
      t2 = 0;
      break;
    case 3:
      t0 = 0;
      t1 = 1;
      t2 = 1;
      break;
  }
}
int main() {
  int n;
  cin >> n;
  int i;
  int v0 = 0, v1 = 1023;
  for (i = 0; i < n; i++) {
    char op[10];
    int v;
    scanf("%s %d", op, &v);
    if (op[0] == '|') {
      v0 |= v;
      v1 |= v;
    } else if (op[0] == '^') {
      v0 ^= v;
      v1 ^= v;
    } else if (op[0] == '&') {
      v0 &= v;
      v1 &= v;
    } else {
      cout << "ERR" << endl;
    }
  }
  int ans0 = 0, ans1 = 0, ans2 = 0;
  for (i = 9; i >= 0; i--) {
    ans0 *= 2;
    ans1 *= 2;
    ans2 *= 2;
    int t0, t1, t2;
    getbit(v0, v1, i, t0, t1, t2);
    ans0 += t0;
    ans1 += t1;
    ans2 += t2;
  }
  cout << 3 << endl;
  cout << "^ " << ans0 << endl;
  cout << "& " << ans1 << endl;
  cout << "| " << ans2 << endl;
  return 0;
}