Problem: You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.

You may write new numbers on the blackboard with the following two operations. 

  * You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard. 
  * You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard. 

Perform a sequence of operations such that at the end the number 1 is on the blackboard.

Input

The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).

Output

Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation. 

  * The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard. 
  * The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard. 

The operation symbol (+ or ^) must be separated from a, b by a whitespace.

You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.

Examples

Input


3


Output


5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9


Input


123


Output


10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<char, pair<long long, long long> > > result;
long long moltiplica(long long a, long long b) {
  if (a == 1) {
    return b;
  } else if (a % 2 == 1) {
    result.push_back(pair<char, pair<long long, long long> >(
        '+', pair<long long, long long>(moltiplica(a - 1, b), b)));
    return a * b;
  } else {
    long long s = moltiplica(a / 2, b);
    result.push_back(pair<char, pair<long long, long long> >(
        '+', pair<long long, long long>(s, s)));
    return a * b;
  }
}
pair<long long, long long> rev(pair<long long, long long> pa) {
  pair<long long, long long> res =
      pair<long long, long long>(pa.second, pa.first);
  return res;
}
pair<long long, long long> mcd(long long x, long long y) {
  if (y == 0) {
    return pair<long long, long long>(1, 0);
  }
  if (y == 1) return pair<long long, long long>(0, 1);
  if (x < y) {
    return rev(mcd(y, x));
  }
  long long q = x / y;
  long long r = x % y;
  pair<long long, long long> res = mcd(r, y);
  res = pair<long long, long long>(res.first, -res.first * q + res.second);
  return res;
}
long long absll(long long x) {
  if (x < 0) return -x;
  return x;
}
int main() {
  long long x;
  cin >> x;
  int e = 0;
  long long molt = 1;
  while (x >> e != 0) e++;
  e--;
  molt <<= e;
  long long y = moltiplica(molt, x);
  result.push_back(pair<char, pair<long long, long long> >(
      '^', pair<long long, long long>(y, x)));
  y ^= x;
  pair<long long, long long> pa;
  pa = mcd(x, y);
  if (pa.first > 0 && pa.second % 2 != 0) {
    pa.first += y;
    pa.second -= x;
  } else if (pa.first < 0 && pa.first % 2 != 0) {
    pa.first -= y;
    pa.second += x;
  }
  result.push_back(pair<char, pair<long long, long long> >(
      '^', pair<long long, long long>(moltiplica(absll(pa.first), x),
                                      moltiplica(absll(pa.second), y))));
  cout << result.size() << endl;
  for (auto row : result) {
    cout << row.second.first << " " << row.first << " " << row.second.second
         << endl;
  }
  return 0;
}