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;
struct op {
  char typ;
  long long a, b;
  op(char _typ, long long _a, long long _b) {
    typ = _typ;
    a = _a;
    b = _b;
  }
};
long long get_rand(long long l, long long r) {
  long long aa = (rand() << 15) ^ (rand());
  return aa % (r - l + 1) + l;
}
vector<op> res, cur;
long long n, m, i, j, k, t, t1, u, v, a, b;
vector<long long> base;
int main() {
  srand(time(NULL));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  base.push_back(n);
  for (t = 1; t <= 200000; t++) {
    cur.clear();
    sort(base.begin(), base.end());
    if (base[0] == 1) break;
    u = get_rand(0, base.size() - 1);
    v = get_rand(0, base.size() - 1);
    u = base[u];
    v = base[v];
    cur.emplace_back('+', u, v);
    u += v;
    if (u >= (long long)1 << 58) continue;
    for (i = base.size() - 1; i >= 0; i--) {
      if (floor(log2(u)) == floor(log2(base[i]))) {
        cur.emplace_back('^', u, base[i]);
        u ^= base[i];
      }
    }
    if (u) {
      for (auto g : cur) res.push_back(g);
      base.push_back(u);
    }
  }
  for (t = 1; t <= 200000; t++) {
    cur.clear();
    sort(base.begin(), base.end());
    if (base[0] == 1) break;
    u = get_rand(0, base.size() - 1);
    v = get_rand(0, base.size() - 1);
    u = base[u];
    v = base[v];
    cur.emplace_back('^', u, v);
    u ^= v;
    if (u >= (long long)1 << 58) continue;
    for (i = base.size() - 1; i >= 0; i--) {
      if (floor(log2(u)) == floor(log2(base[i]))) {
        cur.emplace_back('^', u, base[i]);
        u ^= base[i];
      }
    }
    if (u) {
      for (auto g : cur) res.push_back(g);
      base.push_back(u);
    }
  }
  cout << res.size() << endl;
  for (auto g : res) cout << g.a << ' ' << g.typ << ' ' << g.b << endl;
}