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 = 1e5 + 10;
long long A[maxn], B[maxn], C[maxn];
int m;
char op[maxn];
long long Xor(long long a, long long b) {
  A[++m] = a, B[m] = b, C[m] = a ^ b;
  op[m] = '^';
  return a ^ b;
}
long long Add(long long a, long long b) {
  A[++m] = a, B[m] = b, C[m] = a + b;
  op[m] = '+';
  return a + b;
}
void calc(long long x) {
  if (x == 1) return;
  long long y = x, t = x;
  while (t != (t & -t)) t -= t & -t;
  while ((y & -y) != t) y = Add(y, y);
  long long z = Xor(x, y), p = Add(y, z), q = Xor(x, p), h = Xor(q, Add(y, y));
  for (; h <= y; h = Add(h, h))
    if ((h ^ y) < y) y = Xor(h, y);
  calc(Xor(x, y));
}
int main() {
  int n;
  scanf("%d", &n);
  calc(n);
  printf("%d\n", m);
  for (int i = 1; i <= m; i++) printf("%lld %c %lld\n", A[i], op[i], B[i]);
  return 0;
}