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 int maxn = 1e6 + 100;
constexpr int bits(long long x) { return 63 - __builtin_clzll(x); }
string ts(string second) { return second; }
long long get(long long x, long long y) {
  if (x > y) swap(x, y);
  if (x == 0) return y;
  long long m = bits(y) - bits(x);
  return get(x, y ^ (x << m));
}
vector<string> ans;
set<long long> second;
void find_sum(long long z, long long a, long long b) {
  if (second.count(z)) return;
  second.insert(z);
  ans.push_back(ts(to_string(a)) + " + " + ts(to_string(b)));
}
void find_xor(long long z, long long a, long long b) {
  if (second.count(z)) return;
  second.insert(z);
  ans.push_back(ts(to_string(a)) + " ^ " + ts(to_string(b)));
}
void dial(long long x, long long y) {
  if (x > y) swap(x, y);
  if (x == 0) return;
  long long xx = x;
  for (int i = 0; i < bits(y) - bits(x); i++) {
    xx <<= 1;
    find_sum(xx, xx / 2, xx / 2);
  }
  find_xor(xx ^ y, xx, y);
  dial(x, xx ^ y);
}
void check(long long x, long long cur) {
  long long y = x * cur;
  for (long long i = 2; i <= cur; i <<= 1)
    find_sum(x * i, x * i / 2, x * i / 2);
  long long w = 0;
  for (long long i = 1; i <= cur; i <<= 1)
    if (cur & i) {
      if (w == 0)
        w = x * i;
      else {
        find_sum(w + x * i, w, x * i);
        w += x * i;
      }
    }
  dial(x, y);
}
void go(long long x) {
  ans.clear();
  second = {x};
  while (x > 1) {
    long long cur = 1;
    while (get(x, cur * x) == x) cur++;
    check(x, cur);
    long long xx = get(x, cur * x);
    x = xx;
  }
}
int main() {
  int t = 1;
  while (t--) {
    long long x;
    cin >> x;
    go(x);
    cout << ans.size() << "\n";
    for (auto& x : ans) cout << x << "\n";
  }
  return 0;
}