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 B[2000];
using namespace std;
int main() {
  int n, k, i, j, ans, pos, min;
  int A[2000];
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      if (A[j] - A[i] != (j - i) * k && A[i] + (j - i) * k > 0)
        B[i]++;
      else if (A[i] - A[j] != (i - j) * k && A[i] + (j - i) * k <= 0) {
        B[i] = 1e8;
        break;
      }
    }
  pos = 0;
  min = 1e8;
  for (i = 1; i <= n; i++)
    if (min > B[i]) {
      min = B[i];
      pos = i;
    }
  printf("%d\n", min);
  for (i = 1; i <= n; i++) {
    if (A[i] - A[pos] > (i - pos) * k)
      printf("- %d %d\n", i, A[i] - A[pos] - (i - pos) * k);
    if (A[i] - A[pos] < (i - pos) * k)
      printf("+ %d %d\n", i, (i - pos) * k + A[pos] - A[i]);
  }
  return 0;
}
