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>
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int arr[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int count, min = 1001, mem, ch;
  for (int i = 0; i < n; i++) {
    count = 0, ch = 0;
    for (int j = 0; j < n; j++) {
      if (i == j)
        count++;
      else if (i < j) {
        if (arr[j] == arr[i] + (j - i) * k) count++;
      } else {
        if (arr[i] - (i - j) * k == arr[j]) count++;
        if (arr[i] - (i - j) * k <= 0) ch = 1;
      }
    }
    if (n - count < min && ch == 0) {
      min = n - count;
      mem = i;
    }
  }
  printf("%d\n", min);
  for (int i = 0; i < n; i++) {
    if (i > mem) {
      if (arr[i] > arr[mem] + (i - mem) * k) {
        if (arr[i] - (arr[mem] + (i - mem) * k) != 0)
          printf("- %d %d\n", i + 1, arr[i] - (arr[mem] + (i - mem) * k));
      } else {
        if (arr[mem] + (i - mem) * k - arr[i] != 0)
          printf("+ %d %d\n", i + 1, arr[mem] + (i - mem) * k - arr[i]);
      }
    } else {
      if (arr[i] > arr[mem] - (mem - i) * k) {
        if (arr[i] - (arr[mem] - (mem - i) * k) != 0)
          printf("- %d %d\n", i + 1, arr[i] - (arr[mem] - (mem - i) * k));
      } else {
        if (arr[mem] - (mem - i) * k - arr[i] != 0)
          printf("+ %d %d\n", i + 1, arr[mem] - (mem - i) * k - arr[i]);
      }
    }
  }
  return 0;
}
