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 long long K = 1 << 21;
set<long long> S;
vector<long long> A, B;
vector<bool> T;
long long do_sum(long long a, long long b) {
  if (S.count(a + b)) return a + b;
  assert(S.count(a));
  assert(S.count(b));
  S.insert(a + b);
  A.push_back(a);
  B.push_back(b);
  T.push_back(true);
  return a + b;
}
long long do_xor(long long a, long long b) {
  if (S.count(a ^ b)) return a ^ b;
  assert(S.count(a));
  assert(S.count(b));
  S.insert(a ^ b);
  A.push_back(a);
  B.push_back(b);
  T.push_back(false);
  return a ^ b;
}
bool bad(long long n) {
  if (n % 4 == 3) return true;
  return false;
}
long long unbad(long long n) {
  do_sum(n, n);
  return do_xor(2 * n, n);
}
void make_multiple(long long n, long long k) {
  for (long long t = n * 2; t <= k * n; t <<= 1) {
    do_sum(t / 2, t / 2);
  }
  long long cur = 0;
  for (long long t = 1; t <= k; t <<= 1) {
    if (t & k) {
      cur = do_sum(cur, n * t);
    }
  }
  assert(cur == n * k);
}
int main() {
  long long n;
  scanf("%I64d", &n);
  S.insert(n);
  long long n_backup = n;
  do_xor(n, n);
  if (bad(n)) n = unbad(n);
  long long m = n * 3;
  while (m % K != 1) m += 2 * n;
  make_multiple(n, m / n);
  long long sum_nm = do_sum(n, m);
  long long xor_nm = do_xor(n, m);
  long long two = do_xor(sum_nm, xor_nm);
  make_multiple(2, n / 2);
  do_xor(n - 1, n);
  assert(S.count(1));
  printf("%d\n", (int)T.size());
  for (int i = 0; i < T.size(); i++) {
    if (T[i])
      printf("%I64d + %I64d\n", A[i], B[i]);
    else
      printf("%I64d ^ %I64d\n", A[i], B[i]);
  }
  return 0;
}