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<string> ans;
int64_t phi(int64_t x) {
  int ans = x;
  if (x % 2 == 0) ans /= 2;
  while (x % 2 == 0) x /= 2;
  for (int64_t p = 3; p * p <= x; p += 2) {
    if (x % p == 0) {
      ans = ans * (p - 1) / p;
      x /= p;
      while (x % p == 0) x /= p;
    }
  }
  if (x > 1) ans = ans * (x - 1) / x;
  return ans;
}
int64_t mod_pow(int64_t x, int64_t p, int64_t md) {
  x %= md;
  int64_t a = 1;
  while (p) {
    if (p & 1) a = (a * x) % md;
    x = (x * x) % md;
    p >>= 1;
  }
  return a;
}
int main() {
  ios::sync_with_stdio(false);
  int64_t x;
  cin >> x;
  int64_t p2 = 1, y = x;
  while (p2 * 2 < x) {
    p2 *= 2;
    ans.push_back(to_string(y) + " + " + to_string(y));
    y *= 2;
  }
  ans.push_back(to_string(x) + " ^ " + to_string(y));
  y ^= x;
  int64_t b = mod_pow(y, phi(x) - 1, x), a = b * y / x;
  if (a % 2 == 1) {
    b += x;
    a += y;
  }
  int64_t X = 0, Y = 0;
  for (int64_t i = 1, p = x; i <= a; i *= 2, p *= 2) {
    ans.push_back(to_string(p) + " + " + to_string(p));
    if (i & a) {
      if (X) ans.push_back(to_string(X) + " + " + to_string(p));
      X += p;
    }
  }
  for (int64_t i = 1, p = y; i <= b; i *= 2, p *= 2) {
    ans.push_back(to_string(p) + " + " + to_string(p));
    if (i & b) {
      if (Y) ans.push_back(to_string(Y) + " + " + to_string(p));
      Y += p;
    }
  }
  ans.push_back(to_string(X) + " ^ " + to_string(Y));
  cout << ans.size() << endl;
  for (auto& s : ans) cout << s << endl;
}