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;
const long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
template <class T>
vector<T> ReadVector(int n) {
  vector<T> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  return a;
}
template <class T>
void WriteVector(const vector<T>& a) {
  for (int i = 0; i < ((int)(a).size()); ++i) {
    cout << a[i] << (i + 1 < ((int)(a).size()) ? ' ' : '\n');
  }
}
void Solve() {
  long long x;
  cin >> x;
  vector<long long> a;
  a.push_back(x);
  vector<string> res;
  while (a[0] != 1) {
    long long val = -1;
    for (int i = 0; i < ((int)(a).size()); ++i) {
      for (int j = 0; j <= i; ++j) {
        long long tmp = a[i] + a[j];
        val = tmp;
        for (int k = ((int)(a).size()) - 1; k >= 0; --k) {
          tmp = min(tmp, tmp ^ a[k]);
        }
        if (tmp > 0) {
          res.push_back(to_string(a[i]) + " + " + to_string(a[j]));
          break;
        }
        val = -1;
      }
      if (val > 0) {
        break;
      }
    }
    for (int i = ((int)(a).size()) - 1; i >= 0; --i) {
      if ((val ^ a[i]) < val) {
        res.push_back(to_string(a[i]) + " ^ " + to_string(val));
        val ^= a[i];
      }
    }
    for (int i = 0; i < ((int)(a).size()); ++i) {
      if ((a[i] ^ val) < a[i]) {
        res.push_back(to_string(a[i]) + " ^ " + to_string(val));
        a[i] ^= val;
      }
    }
    a.push_back(val);
    sort((a).begin(), (a).end());
  }
  cout << ((int)(res).size()) << "\n";
  for (int i = 0; i < ((int)(res).size()); ++i) {
    cout << res[i] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  std::cin.tie(nullptr);
  int tc = 1;
  for (int ti = 0; ti < tc; ++ti) {
    Solve();
  }
  return 0;
}