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 q[1005], w[1005];
int main() {
  int n, k;
  int mmx = 10000;
  int i, j;
  int mm = 0;
  memset(q, 0, sizeof(q));
  memset(w, 0, sizeof(w));
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &q[i]);
  }
  for (i = 0; i < n; i++) {
    if ((q[i] - i * k) > 0 && (q[i] + (n - i - 1) * k) > 0) {
      for (j = 0; j < n; j++) {
        if (q[j] != (q[i] + (j - i) * k)) {
          w[i]++;
        }
      }
      if (mmx > w[i]) {
        mmx = w[i];
        mm = i;
      }
    }
  }
  if (w[mm] == 0) {
    printf("0");
  } else {
    printf("%d\n", w[mm]);
    for (i = 0; i < n; i++) {
      int h = q[mm] + (i - mm) * k - q[i];
      if (h > 0) printf("+ %d %d\n", i + 1, h);
      if (h < 0) printf("- %d %d\n", i + 1, -h);
    }
  }
  return 0;
}
