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, k, ans = 10000000, a[1005], v = 0;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    if (a[i] - i * k < 1) continue;
    int p = 0;
    for (int j = 0; j < n; j++) {
      int w;
      if (j > i)
        w = a[i] + (j - i) * k;
      else
        w = a[i] - (i - j) * k;
      if (w != a[j]) p++;
    }
    if (p < ans) ans = p, v = i;
  }
  printf("%d\n", ans);
  for (int i = 0; i < n; i++) {
    int w;
    if (i > v)
      w = a[v] + (i - v) * k;
    else
      w = a[v] - (v - i) * k;
    if (w > a[i]) printf("+ %d %d\n", i + 1, w - a[i]);
    if (w < a[i]) printf("- %d %d\n", i + 1, a[i] - w);
  }
  return 0;
}
