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 {
  int op;
  long long a, b;
};
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, y, x);
  y -= x * (a / b);
  return d;
}
vector<Node> ans;
void solve(long long x, long long k) {
  for (long long i = 1; i <= k; i <<= 1) ans.push_back({0, x * i, x * i});
  long long y = x, val = 0;
  while (k) {
    if (k & 1) {
      ans.push_back({0, val, y});
      val += y;
    }
    y <<= 1;
    k >>= 1;
  }
}
int main() {
  long long x;
  scanf("%lld", &x);
  ans.push_back({1, x, x});
  long long k = 1;
  while (k <= x) k <<= 1;
  k >>= 1;
  solve(x, k);
  long long y = x * k;
  ans.push_back({1, x, y});
  y ^= x;
  long long a, b;
  exgcd(x, y, a, b);
  a = abs(a);
  b = abs(b);
  solve(x, a);
  solve(y, b);
  long long p = x * a, q = y * b;
  if ((p ^ q) != 1) {
    ans.push_back({0, p, x});
    ans.push_back({0, q, x});
    p += x;
    q += x;
  }
  ans.push_back({1, p, q});
  printf("%d\n", ans.size());
  for (auto u : ans) {
    printf("%lld", u.a);
    if (u.op == 0)
      printf(" + ");
    else
      printf(" ^ ");
    printf("%lld\n", u.b);
  }
  return 0;
}