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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
vector<string> calcs;
vector<long long> basis;
bool sim(long long a) {
  long long A = a;
  for (long long b : basis) {
    A = min(A, A ^ b);
  }
  return A;
}
void add(long long a) {
  long long A = a;
  for (long long b : basis) {
    if ((A ^ b) < A) {
      calcs.push_back(to_string(A) + " ^ " + to_string(b));
      A = A ^ b;
    }
  }
  if (A) {
    int ind = 0;
    while (ind < basis.size() && basis[ind] > A) ++ind;
    basis.insert(basis.begin() + ind, A);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
  mt19937 rng(seed);
  long long x;
  cin >> x;
  add(x);
  while (true) {
    long long a = rng() % basis.size();
    long long b = rng() % basis.size();
    if (basis[a] > 1000000000000000000LL || basis[b] > 1000000000000000000LL)
      continue;
    long long x = basis[a] + basis[b];
    if (sim(x)) {
      calcs.push_back(to_string(basis[a]) + " + " + to_string(basis[b]));
      add(x);
    }
    if (basis[basis.size() - 1] == 1) break;
  }
  cout << calcs.size() << "\n";
  for (auto u : calcs) cout << u << "\n";
  return 0;
}