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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void exgcd(long long a, long long &x, long long b, long long &y,
           long long mod) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, y, a % b, x, mod);
  y = (y - a / b * x % mod + mod) % mod;
}
long long ny(long long a, long long mod) {
  long long x, y;
  exgcd(a, x, mod, y, mod);
  return x;
}
long long tot;
struct opt {
  long long ty;
  long long x, y;
  opt(long long ty_ = 0, long long x_ = 0, long long y_ = 0)
      : ty(ty_), x(x_), y(y_) {}
} ans[100010];
void suan(long long m, long long k) {
  long long u = m;
  while (k && !(k & 1)) {
    ans[++tot] = opt(1, u, u);
    u <<= 1;
    k >>= 1;
  }
  long long start = u;
  k >>= 1;
  while (k) {
    ans[++tot] = opt(1, u, u);
    u <<= 1;
    if (k & 1) ans[++tot] = opt(1, start, u), start += u;
    k >>= 1;
  }
}
signed main() {
  long long n;
  cin >> n;
  bool flag = 0;
  long long G;
  for (long long i = 0; !flag && i <= 20; i++)
    for (long long j = 0; !flag && j <= 20; j++) {
      G = (n << i) ^ (n << j);
      if (gcd(G, n) == 1) {
        suan(n, 1 << i);
        suan(n, 1 << j);
        ans[++tot] = opt(0, n << i, n << j);
        flag = 1;
        break;
      }
    }
  long long A = ny(G, n);
  suan(G, A);
  long long X = A * G;
  long long X_1 = X - 1;
  suan(n, X_1 / n);
  if (X_1 % 2 == 0)
    ans[++tot] = opt(0, X_1, X);
  else {
    ans[++tot] = opt(1, X_1, X);
    ans[++tot] = opt(1, X_1, X_1);
    ans[++tot] = opt(0, X_1 << 1, X_1 + X);
  }
  cout << tot << '\n';
  for (long long i = 1; i <= tot; i++) {
    cout << ans[i].x;
    if (ans[i].ty)
      cout << " + ";
    else
      cout << " ^ ";
    cout << ans[i].y << "\n";
  }
  return 0;
}