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>
int tot;
char ans[1000006], tmp[233];
struct Node {
  long long v;
  Node(long long x = 0) : v(x) {}
};
Node operator+(Node a, Node b) {
  ++tot;
  sprintf(tmp, "%lld + %lld\n", a.v, b.v);
  strcat(ans, tmp);
  return Node(a.v + b.v);
}
Node operator<<(Node a, int b) {
  for (int i = 1; i <= b; ++i) {
    ++tot;
    sprintf(tmp, "%lld + %lld\n", a.v, a.v);
    strcat(ans, tmp);
    a.v <<= 1;
  }
  return Node(a.v);
}
Node operator^(Node a, Node b) {
  ++tot;
  sprintf(tmp, "%lld ^ %lld\n", a.v, b.v);
  strcat(ans, tmp);
  return Node(a.v ^ b.v);
}
long long operator&(Node a, Node b) { return a.v & b.v; }
int cal(long long v) {
  for (int x = 0;; ++x, v >>= 1)
    if (!v) return x;
}
int main() {
  int x, m;
  Node a, b, c, d, e;
  scanf("%d", &x);
  a = Node(x);
  if (x % 4 == 1) {
    m = cal(a.v);
    b = a << (m - 1);
    e = (a + b) ^ (a ^ b);
  } else {
    c = a + a;
    d = a ^ c;
    m = cal(d.v);
    b = d << (m - 1);
    e = (d + b) ^ (d ^ b);
  }
  m = cal(e.v);
  for (int i = 1; i < m; ++i) {
    if (a & e) a = a ^ e;
    b = a;
    for (int j = i + 1; j < m; ++j) {
      b = b << 1;
      if (b & e) b = b ^ e;
    }
    e = b;
  }
  printf("%d\n", tot);
  printf("%s", ans);
  return 0;
}