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 k, n, hight[1001], f[1001], ans[1001];
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", hight + i);
    int min = 1001;
    for (int i = 1; i <= 1000; i++) {
      int d = 0;
      int temp1 = i - k;
      for (int j = 1; j <= n; j++) {
        temp1 = temp1 + k;
        f[j] = temp1 - hight[j];
        if (f[j] != 0) d++;
      }
      if (d <= min) {
        min = d;
        for (int k = 1; k <= n; k++) ans[k] = f[k];
      }
    }
    printf("%d\n", min);
    for (int i = 1; i <= n; i++)
      if (ans[i] > 0)
        printf("+ %d %d\n", i, ans[i]);
      else if (ans[i] < 0)
        printf("- %d %d\n", i, -ans[i]);
  }
  return 0;
}
