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;
#pragma GCC optimize("O3")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
template <int, typename T>
struct MINMAX {
  T val;
  MINMAX(T val) : val(val) {}
};
template <typename T>
MINMAX<1, T> MAX(T val) {
  return MINMAX<1, T>(val);
};
template <typename T>
MINMAX<2, T> MIN(T val) {
  return MINMAX<2, T>(val);
};
template <typename T, typename U>
inline T& operator|=(T& lhs, MINMAX<1, U> rhs) {
  return lhs = max(lhs, rhs.val);
}
template <typename T, typename U>
inline T& operator|=(T& lhs, MINMAX<2, U> rhs) {
  return lhs = min(lhs, rhs.val);
}
template <typename T, typename U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T = int>
inline vector<T> READ(int n) {
  vector<T> vec(n);
  for (int i = 0; i < int(n); i++) cin >> vec[i];
  return vec;
}
template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
int log_2(signed long long int n) {
  assert(n > 0);
  return 63 - __builtin_clzll(n);
}
vector<tuple<signed long long int, char, signed long long int>> ops;
signed long long int Add(signed long long int a, signed long long int b) {
  ops.push_back({a, '+', b});
  return a + b;
}
signed long long int Xor(signed long long int a, signed long long int b) {
  ops.push_back({a, '^', b});
  return a ^ b;
}
void solve(signed long long int a) {
  signed long long int b = a;
  for (int i = 0; i < int(log_2(a)); i++) {
    b = Add(b, b);
  }
  signed long long int apb = Add(a, b);
  signed long long int axb = Xor(a, b);
  signed long long int c = Xor(apb, axb);
  assert(log_2(c) == log_2(a) + 1);
  signed long long int d = a;
  while (__builtin_popcount(d) >= 2) {
    d = Add(d, d);
    if (log_2(d) == log_2(c)) d = Xor(d, c);
  }
  while (log_2(d) < log_2(a)) {
    d = Add(d, d);
  }
  d = Xor(d, a);
  if (d > 1) solve(d);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(12);
  signed long long int x;
  cin >> x;
  if ((x & 3) == 3) x = Xor(x, Add(x, x));
  solve(x);
  cout << ops.size() << endl;
  for (auto [a, b, c] : ops) {
    cout << a << " " << b << " " << c << endl;
  }
}