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 n_trees, k_factor;
  while (cin >> n_trees >> k_factor) {
    int model;
    int high[n_trees];
    for (int i = 0; i < n_trees; i++) {
      cin >> high[i];
    }
    int menor = 10000;
    int indice = -1;
    for (int i = 0; i < n_trees; i++) {
      int base = high[i] - i * k_factor;
      int cambio = 0;
      if (base <= 0) continue;
      for (int j = 0; j < n_trees; j++) {
        if (high[j] != base) cambio++;
        base += k_factor;
      }
      if (cambio < menor) {
        menor = cambio;
        indice = i;
      }
    }
    cout << menor << endl;
    int base = high[indice] - indice * k_factor;
    for (int i = 0; i < n_trees; i++) {
      if (base != high[i]) {
        int res = base - high[i];
        if (res >= 0)
          cout << "+ ";
        else
          cout << "- ";
        printf("%d %d \n", i + 1, abs(res));
      }
      base += k_factor;
    }
  }
}
