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;
int x;
vector<tuple<long long, char, long long> > ans;
inline void exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  } else {
    exgcd(b, a % b, y, x);
    y -= a / b * x;
  }
}
inline void pushadd(long long a, long long b) {
  ans.emplace_back(make_tuple(a, '+', b));
}
inline void pushxor(long long a, long long b) {
  ans.emplace_back(make_tuple(a, '^', b));
}
inline void pushmul(long long n, long long x) {
  long long ret = x;
  --n;
  while (n > 0) {
    if (n & 1) pushadd(ret, x), ret += x;
    pushadd(x, x);
    x += x;
    n >>= 1;
  }
}
int main() {
  scanf("%d", &x);
  long long y = 1LL << (63LL - __builtin_clzll(x));
  pushmul(y, x);
  pushxor(x * y, x);
  y = (x * y) ^ x;
  long long a, b;
  exgcd(x, y, a, b);
  b = -b;
  for (long long i = (1LL << 60); i; i >>= 1LL)
    if (a + (__int128)i * y <= 0 || b + (__int128)i * x <= 0)
      a += (__int128)i * y, b += (__int128)i * x;
  a += y;
  b += x;
  if (b & 1) a += y, b += x;
  pushmul(a, x);
  pushmul(b, y);
  pushxor(a * x, b * y);
  printf("%d\n", (int)(ans.size()));
  for (int i = 0; i < (int)(ans.size()); ++i)
    printf("%lld %c %lld\n", get<0>(ans[i]), get<1>(ans[i]), get<2>(ans[i]));
  return 0;
}