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 main() {
  int kbeer = (int)1e9;
  int n, k, scnt = 0;
  cin >> n >> k;
  vector<int> list(n);
  vector<pair<int, int> > my(n);
  vector<pair<int, int> > h7otfeeh;
  for (int i = 0; i < n; i++) cin >> list[i];
  for (int i = 1; i <= 1000; i++) {
    int l = i;
    int count = 0;
    my.clear();
    vector<pair<int, int> > my(n);
    for (int j = 0; j < n; j++) {
      if (l != list[j]) {
        count++;
        my[j].first = j + 1;
        my[j].second = list[j] - l;
      }
      l += k;
    }
    if (kbeer > count) {
      h7otfeeh = my;
      scnt = count;
      kbeer = count;
    }
  }
  cout << scnt << endl;
  for (int i = 0; i < n; i++) {
    if (h7otfeeh[i].second != 0) {
      if (h7otfeeh[i].second > 0)
        cout << "-";
      else
        cout << "+";
      cout << " " << h7otfeeh[i].first << " ";
      cout << abs(h7otfeeh[i].second) << endl;
    }
  }
}
