The Queen of England has n trees growing in a row in her garden. At that, the i-th (1 ≤ i ≤ n) tree from the left has height ai meters. Today the Queen decided to update the scenery of her garden. She wants the trees' heights to meet the condition: for all i (1 ≤ i < n), ai + 1 - ai = k, where k is the number the Queen chose.

Unfortunately, the royal gardener is not a machine and he cannot fulfill the desire of the Queen instantly! In one minute, the gardener can either decrease the height of a tree to any positive integer height or increase the height of a tree to any positive integer height. How should the royal gardener act to fulfill a whim of Her Majesty in the minimum number of minutes?

Input

The first line contains two space-separated integers: n, k (1 ≤ n, k ≤ 1000). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the heights of the trees in the row. 

Output

In the first line print a single integer p — the minimum number of minutes the gardener needs. In the next p lines print the description of his actions. 

If the gardener needs to increase the height of the j-th (1 ≤ j ≤ n) tree from the left by x (x ≥ 1) meters, then print in the corresponding line "+ j x". If the gardener needs to decrease the height of the j-th (1 ≤ j ≤ n) tree from the left by x (x ≥ 1) meters, print on the corresponding line "- j x".

If there are multiple ways to make a row of trees beautiful in the minimum number of actions, you are allowed to print any of them.

Examples

Input

4 1
1 2 1 5


Output

2
+ 3 2
- 4 1


Input

4 1
1 2 3 4


Output

0
#include <bits/stdc++.h>
using namespace std;
int a[1004], b[1004];
vector<pair<char, pair<int, int> > > v, sol;
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    sol.push_back(make_pair('-', make_pair(1, 1)));
  }
  sol.push_back(make_pair('-', make_pair(1, 1)));
  for (b[1] = 1; b[1] <= 1000; ++b[1]) {
    v.clear();
    int x = b[1];
    if (a[1] > x)
      v.push_back(make_pair('-', make_pair(1, a[1] - x)));
    else if (a[1] != x)
      v.push_back(make_pair('+', make_pair(1, x - a[1])));
    for (int i = 2; i <= n; ++i) {
      int x = b[i - 1] + k;
      if (a[i] > x)
        v.push_back(make_pair('-', make_pair(i, a[i] - x)));
      else if (a[i] != x)
        v.push_back(make_pair('+', make_pair(i, x - a[i])));
      b[i] = x;
    }
    if (sol.size() > v.size()) {
      sol.clear();
      for (unsigned j = 0; j < v.size(); ++j) sol.push_back(v[j]);
    }
  }
  cout << sol.size() << "\n";
  for (unsigned i = 0; i < sol.size(); ++i)
    cout << sol[i].first << " " << sol[i].second.first << " "
         << sol[i].second.second << "\n";
  return 0;
}
