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;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int mn = 52 + 10;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
const long long half = (mod + 1) / 2;
const double PI = acos(-1);
vector<string> v;
void ad(long long a, long long b) {
  v.push_back(to_string(a) + " + " + to_string(b));
}
void xo(long long a, long long b) {
  v.push_back(to_string(a) + " ^ " + to_string(b));
}
void solve(long long x) {
  long long lp = 1LL << __builtin_ctz(x);
  if (x == lp) return;
  int a = 0, b = 0;
  for (int i = 0; i < 20; i++)
    if ((x >> i) & 1) {
      b = a;
      a = i;
    }
  int d = a - b;
  for (int i = 0; i < d; i++) ad(x << i, x << i);
  long long xd = x << d;
  xo(xd, x);
  long long cs = (x << d) ^ x;
  ad(cs, x), ad(cs, x << 1), ad(xd, x), ad(xd, x << 1);
  long long csp = cs + x, cspp = cs + x * 2;
  long long xdp = xd + x, xdpp = xd + x * 2;
  xo(csp, xd);
  xo(csp, xdp);
  xo(cspp, xd);
  xo(cspp, xdp);
  xo(cspp, xdpp);
  long long nx = min({csp ^ xd, csp ^ xdp, cspp ^ xd, cspp ^ xdp, cspp ^ xdpp});
  solve(nx);
  long long pp = 1LL << __builtin_ctz(nx);
  if (pp == lp) return;
  for (long long i = pp; x > i; i <<= 1) {
    if (x & i) {
      xo(x, i);
      x ^= i;
    }
    ad(i, i);
  }
  if (x != lp) solve(x);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  long long x;
  cin >> x;
  solve(x);
  printf("%d\n", v.size());
  for (string s : v) printf("%s\n", s.c_str());
}