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;
template <typename T>
struct outputer;
struct outputable {};
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
struct Input {
  int64_t n;
  bool read() { return !!(cin >> n); }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  vector<tuple<int64_t, char, int64_t>> ans;
  void write() {
    cout << ((int)(ans).size()) << endl;
    for (auto t : ans) {
      printf("%" PRId64 " %c %" PRId64 "\n", get<0>(t), get<1>(t), get<2>(t));
      if (get<1>(t) == '+') {
        ;
      } else {
        ;
      }
    }
  }
};
namespace Main {
struct Solution : Data {
  int get_s(int64_t x) {
    int k = 0;
    while (x > 0) {
      x /= 2;
      k++;
    }
    return k;
  }
  void add(int64_t x, int k) {
    int64_t sum = 0;
    while (k) {
      if (k & 1) {
        ans.emplace_back(sum, '+', x);
        sum += x;
      }
      ans.emplace_back(x, '+', x);
      x += x;
      k >>= 1;
    }
  }
  int64_t a[100];
  void solve() {
    ans.emplace_back(n, '^', n);
    memset(a, 0, sizeof a);
    for (int i = int(1); i < int(1e9); ++i) {
      int prev_sz = ((int)(ans).size());
      add(n, i);
      int64_t val = n * i;
      int64_t* ptr;
      while (*(ptr = &a[get_s(val)]) != 0) {
        ans.emplace_back(val, '^', *ptr);
        val ^= *ptr;
      }
      if (val == 0) {
        ans.resize(prev_sz);
      }
      *ptr = val;
      if (val == 1) {
        return;
      }
    }
  }
  void clear() { *this = Solution(); }
};
}  // namespace Main
Main::Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}