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 node {
  long long tp, x, y;
};
vector<node> ans;
void Mul(long long x, long long y) {
  long long s = 0;
  for (; y;) {
    if (y & 1) ans.push_back((node){2, s, x}), s += x;
    if (y /= 2) ans.push_back((node){2, x, x}), x += x;
  }
}
void exgcd(long long x, long long y, long long &a, long long &b) {
  if (!y) {
    a = 1;
    b = 0;
    return;
  }
  exgcd(y, x % y, b, a);
  b -= x / y * a;
}
int main() {
  long long x;
  scanf("%lld", &x);
  ans.push_back((node){1, x, x});
  long long y = 1;
  for (; y * 2 <= x; y *= 2)
    ;
  Mul(x, y);
  long long z = x * y;
  ans.push_back((node){1, z, x});
  z ^= x;
  long long a, b;
  exgcd(x, z, a, b);
  b *= -1;
  for (; a < 0 || b < 0; b += x, a += z)
    ;
  if (b % 2 == 1) b += x, a += z;
  Mul(x, a);
  Mul(z, b);
  ans.push_back((node){1, x * a, z * b});
  printf("%d\n", ans.size());
  for (auto i : ans) printf("%lld %c %lld\n", i.x, i.tp == 1 ? '^' : '+', i.y);
}