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 <class T>
void print(T& x) {
  cout << x;
}
template <class T, class U>
void print(pair<T, U>& p) {
  cout << "(";
  print(p.first);
  cout << ", ";
  print(p.second);
  cout << ")";
}
template <class T>
void print(vector<T>& v) {
  cout << "{";
  if (((int)(v).size())) print(v[0]);
  for (int i = 1; i < ((int)(v).size()); i++) cout << ", ", print(v[i]);
  cout << "}\n";
}
vector<long long> X;
vector<tuple<int, long long, long long>> Q;
long long add(long long x) {
  for (long long z : X) {
    if ((z ^ x) < x) {
      Q.emplace_back(1, x, z);
      x = min(x, (z ^ x));
    }
  }
  if (x > 0 && x < 1e17) {
    for (int i = 0; i < ((int)(X).size()); i++) {
      if (X[i] > (X[i] ^ x)) {
        Q.emplace_back(1, X[i], x);
        X[i] = min(X[i], X[i] ^ x);
      }
    }
    X.push_back(x);
  } else
    x = 0;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long x;
  cin >> x;
  queue<long long> q;
  q.push(x);
  while (((int)(q).size())) {
    long long res = q.front();
    q.pop();
    res = add(res);
    if (res) {
      for (long long k : X) {
        Q.emplace_back(0, k, res);
        q.push(k + res);
      }
    }
    if (res == 1) {
      break;
    }
  }
  cout << ((int)(Q).size()) << endl;
  for (auto [t, a, b] : Q) {
    cout << a;
    if (t == 0) {
      cout << " + ";
    } else
      cout << " ^ ";
    cout << b << "\n";
  }
}