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 arr[10001], brr[10001], crr[10001];
using namespace std;
int main() {
  int i, j, pos, count, k, n, mini_moves;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; ++i) scanf("%d", arr + i);
  mini_moves = 100000;
  count = 0;
  pos = 0;
  for (i = 0; i < n; ++i) {
    brr[i] = arr[i];
    for (j = i - 1; j >= 0; j--) brr[j] = brr[j + 1] - k;
    for (j = i + 1; j < n; ++j) brr[j] = brr[j - 1] + k;
    for (j = 0; j < n; ++j)
      if (brr[j] <= 0) break;
    if (j != n) continue;
    for (j = 0; j < n; ++j) crr[j] = arr[j] - brr[j];
    count = 0;
    for (j = 0; j < n; ++j) {
      if (crr[j] != 0) count++;
    }
    if (count < mini_moves) {
      mini_moves = count;
      pos = i;
    }
  }
  brr[pos] = arr[pos];
  for (j = pos + 1; j < n; ++j) brr[j] = brr[j - 1] + k;
  for (j = pos - 1; j >= 0; j--) {
    brr[j] = brr[j + 1] - k;
  }
  for (j = 0; j < n; ++j) crr[j] = arr[j] - brr[j];
  printf("%d\n", mini_moves);
  for (j = 0; j < n; ++j) {
    if (crr[j] != 0) {
      if (crr[j] > 0) {
        printf("- %d %d\n", j + 1, crr[j]);
      } else
        printf("+ %d %d\n", j + 1, -1 * crr[j]);
    }
  }
  return 0;
}
