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;
int hibit(long long x) {
  for (int i = 63; i >= 0; i--)
    if (x & (1LL << i)) return i;
  assert(false);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  vector<pair<int, long long>> base;
  int len = hibit(x);
  vector<pair<pair<long long, long long>, char>> sol_second_part;
  vector<int> need;
  for (int i = 2;; i += 2) {
    long long y = (long long)i * x;
    for (auto it : base) {
      if (y & (1LL << it.first)) y ^= it.second;
    }
    if (y != 0) {
      y = (long long)i * x;
      for (auto it : base) {
        if (y & (1LL << it.first)) {
          sol_second_part.push_back({{y, it.second}, '^'});
          y ^= it.second;
        }
      }
      base.push_back({hibit(y), y});
      len = max(len, hibit(y));
      need.push_back(i);
      if (len == (int)base.size()) break;
    }
  }
  assert(len == (int)base.size());
  long long y = x;
  for (auto it : base) {
    if (y & (1LL << it.first)) {
      sol_second_part.push_back({{y, it.second}, '^'});
      y ^= it.second;
    }
  }
  assert(y == 1);
  vector<pair<pair<long long, long long>, char>> sol;
  for (int i = 2; i <= need.back(); i *= 2) {
    sol.push_back({{x * i / 2, x * i / 2}, '+'});
  }
  for (auto it : need) {
    int cur = 0;
    for (int i = 0; (1 << i) <= it; i++) {
      if (!((1 << i) & it)) continue;
      if (cur != 0) sol.push_back({{x * (1 << i), x * cur}, '+'});
      cur ^= 1 << i;
    }
    assert(cur == it);
  }
  for (auto it : sol_second_part) sol.push_back(it);
  assert(sol.size() <= 100000);
  cout << sol.size() << '\n';
  for (auto it : sol) {
    cout << it.first.first << ' ' << it.second << ' ' << it.first.second
         << '\n';
  }
  return 0;
}