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(void) {
  int arr[1020];
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int mini = 1000000000;
  int remember = -1;
  for (int i = 0; i < n; i++) {
    if (arr[i] - k * (i) > 0) {
      int coba = 0;
      for (int j = 0; j < n; j++) {
        if (arr[j] != arr[i] - k * (i - j)) coba++;
      }
      if (mini > coba) {
        mini = coba;
        remember = i;
      }
    }
  }
  printf("%d\n", mini);
  for (int i = 0; i < n; i++) {
    int seharusnya = arr[remember] - k * (remember - i);
    if (arr[i] < seharusnya) {
      printf("+ %d %d\n", i + 1, seharusnya - arr[i]);
    } else if (arr[i] > seharusnya) {
      printf("- %d %d\n", i + 1, arr[i] - seharusnya);
    }
  }
  return 0;
}
