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 + 10;
char s[maxn];
int num[maxn];
int one[20], zero[20];
vector<int> ans[5];
int cal(int t, int n) {
  for (int i = 0; i < n; i++) {
    if (s[i] == '|') t |= num[i];
    if (s[i] == '&') t &= num[i];
    if (s[i] == '^') t ^= num[i];
  }
  return t;
}
void Print() {
  int cnt = 0, ans0, ans1, ans3;
  ans0 = ans1 = ans3 = -1;
  for (int i = 0; i < 4; i++)
    if (ans[i].size()) {
      if (i == 2) continue;
      int temp;
      if (i == 0) {
        cnt++;
        temp = 1023;
        for (int j = 0; j < ans[i].size(); j++) {
          temp -= (1 << ans[i][j]);
        }
        ans0 = temp;
      }
      if (i == 1) {
        cnt++;
        temp = 1023;
        for (int j = 0; j < ans[i].size(); j++) {
          temp -= (1 << ans[i][j]);
        }
        temp ^= 1023;
        ans1 = temp;
      }
      if (i == 3) {
        cnt++;
        temp = 0;
        for (int j = 0; j < ans[i].size(); j++) {
          temp += (1 << ans[i][j]);
        }
        ans3 = temp;
      }
    }
  printf("%d\n", cnt);
  if (ans0 != -1) printf("& %d\n", ans0);
  if (ans1 != -1) printf("^ %d\n", ans1);
  if (ans3 != -1) printf("| %d\n", ans3);
}
int main() {
  int n;
  char tt[3];
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s%d", tt, &num[i]);
    s[i] = tt[0];
  }
  int t = 1023, p;
  p = cal(t, n);
  for (int i = 0; i < 10; i++) {
    one[i] = p % 2;
    p /= 2;
  }
  t = 0;
  p = cal(t, n);
  for (int i = 0; i < 10; i++) {
    zero[i] = p % 2;
    p /= 2;
  }
  for (int i = 0; i < 10; i++) {
    if (one[i] == 0 && zero[i] == 0) ans[0].push_back(i);
    if (one[i] == 0 && zero[i] == 1) ans[1].push_back(i);
    if (one[i] == 1 && zero[i] == 0) ans[2].push_back(i);
    if (one[i] == 1 && zero[i] == 1) ans[3].push_back(i);
  }
  Print();
  return 0;
}
