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;
long long n;
struct op {
  long long x, y;
  char c;
};
vector<op> ans;
int get_e(long long x) {
  int ans = 0;
  while (x > 1) {
    ans++;
    x >>= 1;
  }
  return ans;
}
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 tx, ty, c = exgcd(b, a % b, tx, ty);
  x = -ty;
  y = -(tx + a / b * ty);
  return c;
}
void wk(long long x, long long p) {
  long long s = 0, now = x;
  while (p) {
    if (p & 1) {
      if (s) ans.push_back({s, now, '+'});
      s += now;
    }
    p >>= 1;
    ans.push_back({now, now, '+'});
    now <<= 1;
  }
}
int main() {
  scanf("%lld", &n);
  int e = get_e(n);
  long long now = n;
  for (int i = 0; i < e; i++) {
    ans.push_back({now, now, '+'});
    now <<= 1;
  }
  ans.push_back({now, n, '^'});
  now ^= n;
  long long x, y;
  exgcd(now, n, x, y);
  if (x < 0) {
    long long hh = -x / n;
    x += (hh + 1) * n;
    y += (hh + 1) * now;
  }
  if (y < 0) {
    long long hh = -y / now;
    x += (hh + 1) * n;
    y += (hh + 1) * now;
  }
  long long hh = min(x / n, y / now);
  x -= hh * n;
  y -= hh * now;
  if (y & 1) {
    y += now;
    x += n;
  }
  wk(now, x);
  wk(n, y);
  ans.push_back({now * x, n * y, '^'});
  printf("%d\n", ans.size());
  for (op x : ans) printf("%lld %c %lld\n", x.x, x.c, x.y);
}