#include <bits/stdc++.h>
const double pi = acos(-1.0);
using namespace std;
char a[500010];
int b[500010];
int yu[20], huo[20], yihuo[20];
int main() {
  int n;
  cin >> n;
  int c = 0, d = 1023;
  for (int i = 1; i <= n; i++) {
    getchar();
    scanf("%c%d", &a[i], &b[i]);
    if (a[i] == '&') {
      c &= b[i];
      d &= b[i];
    }
    if (a[i] == '|') {
      c |= b[i];
      d |= b[i];
    }
    if (a[i] == '^') {
      c ^= b[i];
      d ^= b[i];
    }
  }
  int cnt = 1;
  for (int i = 1; i <= 10; i++) {
    int cc = c % 2;
    int dd = d % 2;
    if (cc == 0 && dd == 0) {
      yu[cnt] = 0;
      huo[cnt] = 0;
      yihuo[cnt] = 0;
    }
    if (cc == 0 && dd == 1) {
      yu[cnt] = 1;
      huo[cnt] = 0;
      yihuo[cnt] = 0;
    }
    if (cc == 1 && dd == 1) {
      yu[cnt] = 1;
      huo[cnt] = 1;
      yihuo[cnt] = 0;
    }
    if (cc == 1 && dd == 0) {
      yu[cnt] = 1;
      huo[cnt] = 0;
      yihuo[cnt] = 1;
    }
    cnt++;
    c /= 2;
    d /= 2;
  }
  int ans1 = 0, ans2 = 0, ans3 = 0;
  for (int i = 0; i <= 9; i++) {
    ans1 += (pow(2, i) * yu[i + 1]);
    ans2 += (pow(2, i) * huo[i + 1]);
    ans3 += (pow(2, i) * yihuo[i + 1]);
  }
  cout << "3" << endl;
  cout << "& " << ans1 << endl;
  cout << "| " << ans2 << endl;
  cout << "^ " << ans3 << endl;
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
const double pi = acos(-1.0);
using namespace std;
char a[500010];
int b[500010];
int yu[20], huo[20], yihuo[20];
int main() {
  int n;
  cin >> n;
  int c = 0, d = 1023;
  for (int i = 1; i <= n; i++) {
    getchar();
    scanf("%c%d", &a[i], &b[i]);
    if (a[i] == '&') {
      c &= b[i];
      d &= b[i];
    }
    if (a[i] == '|') {
      c |= b[i];
      d |= b[i];
    }
    if (a[i] == '^') {
      c ^= b[i];
      d ^= b[i];
    }
  }
  int cnt = 1;
  for (int i = 1; i <= 10; i++) {
    int cc = c % 2;
    int dd = d % 2;
    if (cc == 0 && dd == 0) {
      yu[cnt] = 0;
      huo[cnt] = 0;
      yihuo[cnt] = 0;
    }
    if (cc == 0 && dd == 1) {
      yu[cnt] = 1;
      huo[cnt] = 0;
      yihuo[cnt] = 0;
    }
    if (cc == 1 && dd == 1) {
      yu[cnt] = 1;
      huo[cnt] = 1;
      yihuo[cnt] = 0;
    }
    if (cc == 1 && dd == 0) {
      yu[cnt] = 1;
      huo[cnt] = 0;
      yihuo[cnt] = 1;
    }
    cnt++;
    c /= 2;
    d /= 2;
  }
  int ans1 = 0, ans2 = 0, ans3 = 0;
  for (int i = 0; i <= 9; i++) {
    ans1 += (pow(2, i) * yu[i + 1]);
    ans2 += (pow(2, i) * huo[i + 1]);
    ans3 += (pow(2, i) * yihuo[i + 1]);
  }
  cout << "3" << endl;
  cout << "& " << ans1 << endl;
  cout << "| " << ans2 << endl;
  cout << "^ " << ans3 << endl;
  return 0;
}

```