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.
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 5;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
template <class T>
inline bool read(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
int cur[10];
int ans[3] = {1023, 0, 0};
int main() {
  int n;
  scanf("%d%*c", &n);
  memset(cur, -1, sizeof(cur));
  while (n--) {
    char ch;
    int x;
    scanf("%c %d%*c", &ch, &x);
    for (int i = 0; i < 10; ++i) {
      int dig = x & 1;
      x >>= 1;
      if (ch == '&' && dig == 0)
        cur[i] = 0;
      else if (ch == '|' && dig == 1)
        cur[i] = 1;
      else if (ch == '^' && dig == 1) {
        if (cur[i] >= 0)
          cur[i] ^= 1;
        else if (cur[i] == -1)
          cur[i] = -2;
        else
          cur[i] = -1;
      }
    }
  }
  for (int i = 0; i < 10; ++i) {
    if (cur[i] == 0)
      ans[0] -= 1 << i;
    else if (cur[i] == 1)
      ans[1] += 1 << i;
    else if (cur[i] == -2)
      ans[2] += 1 << i;
  }
  int cnt = 0;
  if (ans[0] != 1023) cnt++;
  if (ans[1]) cnt++;
  if (ans[2]) cnt++;
  printf("%d\n", cnt);
  if (ans[0] != 1023) printf("& %d\n", ans[0]);
  if (ans[1]) printf("| %d\n", ans[1]);
  if (ans[2]) printf("^ %d\n", ans[2]);
  return 0;
}
