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, p;
  int tree[1001];
  int i, j, b, sum, f, w = 1;
  scanf("%d%d", &n, &k);
  p = n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &tree[i]);
  }
  for (i = 1; i <= n; i++) {
    b = tree[i];
    sum = 0;
    if (b - (i - 1) * k <= 0) continue;
    f = b - (i - 1) * k;
    for (j = 1; j <= i; j++) {
      if (tree[j] != f + (j - 1) * k) sum++;
    }
    for (j = i + 1; j <= n; j++) {
      if (tree[j] != b + (j - i) * k) sum++;
    }
    if (sum < p) {
      p = sum;
      w = i;
    }
  }
  printf("%d\n", p);
  b = tree[w];
  f = b - (w - 1) * k;
  for (j = 1; j <= w; j++) {
    if (tree[j] != f + (j - 1) * k)
      if (tree[j] < f + (j - 1) * k)
        printf("+ %d %d\n", j, f + (j - 1) * k - tree[j]);
      else
        printf("- %d %d\n", j, tree[j] - f - (j - 1) * k);
  }
  for (j = w + 1; j <= n; j++) {
    if (tree[j] != b + (j - w) * k)
      if (tree[j] < b + (j - w) * k)
        printf("+ %d %d\n", j, b + (j - w) * k - tree[j]);
      else
        printf("- %d %d\n", j, tree[j] - b - (j - w) * k);
  }
  return 0;
}
