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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
long long n, m, p;
vector<tuple<long long, char, long long> > ans;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (__int128)ans * a % p;
    a = (__int128)a * a % p;
    b >>= 1;
  }
  return ans;
}
void Get(long long k, long long x) {
  long long now = 0;
  while (k) {
    if (k & 1) {
      if (now) {
        ans.emplace_back(now, '+', x);
      }
      now += x;
    }
    ans.emplace_back(x, '+', x);
    x <<= 1;
    k >>= 1;
  }
}
int main() {
  n = read();
  while ((1LL << (m + 1)) <= n) ++m;
  p = n;
  for (int i = 1; i <= m; ++i) {
    ans.emplace_back(p, '+', p);
    p <<= 1;
  }
  ans.emplace_back(p, '^', n);
  p ^= n;
  long long phi = p, tp = p;
  for (long long i = 2; i * i <= tp; ++i) {
    if (tp % i == 0) {
      phi = phi / i * (i - 1);
      while (tp % i == 0) tp /= i;
    }
  }
  if (tp > 1) phi = phi / tp * (tp - 1);
  long long k = qpow(n, phi - 1) % p;
  assert(k * n % p == 1);
  if (!(k & 1)) k += p;
  Get(k, n);
  Get((k * n - 1) / p, p);
  ans.emplace_back(k * n, '^', k * n - 1);
  printf("%d\n", (int)ans.size());
  for (auto [x, c, y] : ans) {
    printf("%lld %c %lld\n", x, c, y);
  }
  return 0;
}