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;
void draganov_style() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
}
const int NMAX = 5e5 + 4;
int n, vers[14];
int dg[NMAX];
char op[NMAX];
int func(int bit, int t) {
  int x = (t << bit);
  for (int i = 0; i < n; i++) {
    if (op[i] == '|') x |= dg[i];
    if (op[i] == '^') x ^= dg[i];
    if (op[i] == '&') x &= dg[i];
  }
  return ((x >> bit) & 1);
}
int AND = 1023, XOR = 0;
int AND2 = 1023, XOR2 = 0;
void andf(int bit) { AND &= (~(1 << bit)); }
void xrf(int bit) { XOR |= (1 << bit); }
void andf2(int bit) { AND2 &= (~(1 << bit)); }
void xrf2(int bit) { XOR2 |= (1 << bit); }
int main() {
  draganov_style();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> op[i] >> dg[i];
  }
  for (int i = 0; i < 10; i++) {
    int nulBt = func(i, 0);
    int oneBt = func(i, 1);
    vers[i] = ((nulBt << 1) | oneBt);
  }
  for (int i = 0; i < 10; i++) {
    if (vers[i] == 1) continue;
    if (vers[i] == 0) andf(i);
    if (vers[i] == 2) xrf(i);
    if (vers[i] == 3) andf2(i), xrf2(i);
  }
  cout << "4\n& " << AND << "\n^ " << XOR << "\n& " << AND2 << "\n^ " << XOR2;
  return 0;
}