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, a[1001];
  while (scanf("%d%d", &n, &k) != EOF) {
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int s = 1001, index = -1;
    for (int i = 0; i < n; i++) {
      int pre = a[i];
      int cur = 0;
      int flag = 0;
      for (int j = i - 1; j >= 0; j--) {
        if (a[j] != pre - k) cur++;
        pre -= k;
        if (pre <= 0) flag = 1;
      }
      if (flag) continue;
      pre = a[i];
      for (int j = i + 1; j < n; j++) {
        if (a[j] != pre + k) cur++;
        pre += k;
      }
      if (cur < s) {
        index = i;
        s = cur;
      }
    }
    printf("%d\n", s);
    for (int i = 0; i < n; i++) {
      if (i < index) {
        int c = a[index] - (index - i) * k;
        if (a[i] < c)
          printf("+ %d %d\n", i + 1, c - a[i]);
        else if (a[i] > c)
          printf("- %d %d\n", i + 1, a[i] - c);
      } else if (i > index) {
        int c = a[index] + (i - index) * k;
        if (a[i] < c)
          printf("+ %d %d\n", i + 1, c - a[i]);
        else if (a[i] > c)
          printf("- %d %d\n", i + 1, a[i] - c);
      }
    }
  }
  return 0;
}
