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;
struct pack {
  long long op, x, y;
};
vector<pack> ans;
void solve(long long x) {
  if (x == 1) return;
  long long up = 0, x_ = x;
  while (x_) x_ >>= 1, up++;
  long long nw = x;
  for (long long i = 1; i < up; i++) ans.push_back((pack){1, nw, nw}), nw <<= 1;
  long long y = nw;
  ans.push_back((pack){0, nw, x}), nw ^= x;
  ans.push_back((pack){1, nw, y}), nw += y;
  ans.push_back((pack){0, nw, x}), nw ^= x;
  ans.push_back((pack){1, y, y}), y <<= 1;
  ans.push_back((pack){0, nw, y}), nw ^= y;
  assert(nw == (nw & -nw));
  while ((1 << up) <= nw) {
    x_ = x;
    while (x_ != (x_ & -x_)) {
      ans.push_back((pack){1, x_, x_});
      x_ <<= 1;
      if (x_ & nw) {
        ans.push_back((pack){0, x_, nw});
        x_ ^= nw;
      }
    }
    nw = x_;
    assert(nw == (nw & -nw));
  }
  while (nw < (1 << up - 1)) ans.push_back((pack){1, nw, nw}), nw <<= 1;
  ans.push_back((pack){0, x, nw}), x ^= nw, solve(x);
}
signed main() {
  long long x;
  scanf("%lld", &x), solve(x), printf("%lld\n", (long long)ans.size());
  for (auto i : ans)
    printf("%lld ", i.x), i.op ? putchar('+') : putchar('^'),
        printf(" %lld\n", i.y);
  assert(ans.size() <= 100000);
  set<long long> st;
  st.insert(x);
  for (auto i : ans) {
    assert(st.count(i.x) && st.count(i.y));
    i.op ? (st.insert(i.x + i.y), 0) : (st.insert(i.x ^ i.y), 0);
  }
  assert(st.count(1));
  return 0;
}