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;
unordered_map<long long, int> mp;
long long st[100005], ln = 0;
vector<tuple<long long, long long, int> > tp;
void gen(long long x, long long y, int k) {
  if (mp.count(1)) return;
  long long z;
  if (k)
    z = x + y;
  else
    z = x ^ y;
  if (z > 5000000000000000000ll) return;
  if (mp.count(z)) return;
  mp[z] = true;
  st[++ln] = z;
  tp.push_back(make_tuple(x, y, k));
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  } else {
    long long r = exgcd(b, a % b, y, x);
    y -= x * (a / b);
    return r;
  }
}
void get(long long n, long long x) {
  long long res = 0, Pow = x;
  while (n) {
    if (n & 1) {
      gen(res, Pow, 1);
      res += Pow;
    }
    gen(Pow, Pow, 1);
    Pow *= 2;
    n >>= 1;
  }
}
void solve(long long a, long long b) {
  long long x, y;
  exgcd(a, -b, x, y);
  while (min(a * x, b * y) % 2 == 1) y += a, x += b;
  gen(a, a, 0);
  get(x, a);
  get(y, b);
  gen(a * x, b * y, 0);
}
void print() {
  printf("%d\n", tp.size());
  for (auto x : tp) {
    printf("%lld ", get<0>(x));
    if (get<2>(x) == 1)
      printf("+ ");
    else
      printf("^ ");
    printf("%lld\n", get<1>(x));
  }
}
int main() {
  srand((unsigned)time(NULL));
  long long x;
  scanf("%lld", &x);
  st[++ln] = x;
  mp[x] = 1;
  long long y = 1;
  while (y <= x) y *= 2;
  y /= 2;
  get(y, x);
  gen(x, x * y, 0);
  solve(x, x ^ (x * y));
  print();
}