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 gi() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long N = 1e5 + 5;
long long n, top;
struct node {
  long long x, y, z;
  node(long long _x = 0, long long _y = 0, long long _z = 0) {
    x = _x, y = _y, z = _z;
  }
} opt[N];
void f(long long x, long long y, long long z) { opt[++top] = node(x, y, z); }
signed main() {
  n = gi();
  long long x = n;
  for (register long long i = 20; i >= 1; --i) {
    if (!((1 << i) & x)) continue;
    long long y = x;
    for (register long long j = (1); j <= (i); ++j) f(y, y, 1), y <<= 1;
    long long by = y;
    long long z = y ^ x;
    f(x, y, 0);
    long long r = z + y;
    f(z, y, 1);
    long long u = r ^ x;
    f(r, x, 0);
    long long t = y + y;
    f(y, y, 1);
    long long w = t ^ u;
    f(t, u, 0);
    while (by >= w) {
      if (by & w) f(by, w, 0), by ^= w;
      f(w, w, 1), w <<= 1;
    }
    f(by, x, 0), x ^= by;
  }
  cout << top << endl;
  for (register long long i = (1); i <= (top); ++i) {
    if (opt[i].z)
      printf("%lld + %lld\n", opt[i].x, opt[i].y);
    else
      printf("%lld ^ %lld\n", opt[i].x, opt[i].y);
  }
  return 0;
}