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;
  cin >> n >> k;
  int a[1009], mn = 1000, b[1009];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  int ix = 0;
  for (int i = 0; i < n; i++) {
    int m = 0;
    if (a[i] - k * i > 0) {
      for (int j = 0; j < n; j++) {
        if (j < i) {
          if (a[i] - a[j] != k * (i - j)) {
            m++;
          }
        } else {
          if (j > i)
            if (a[j] - a[i] != k * (j - i)) {
              m++;
            }
        }
      }
      if (m <= mn) ix = i, mn = m;
    }
  }
  cout << mn << endl;
  for (int i = ix + 1; i < n; i++) {
    b[i] = b[i - 1] + k;
  }
  for (int i = ix - 1; i > -1; i--) {
    b[i] = b[i + 1] - k;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] < b[i])
      cout << "+ " << i + 1 << ' ' << b[i] - a[i] << endl;
    else if (a[i] > b[i])
      cout << "- " << i + 1 << ' ' << a[i] - b[i] << endl;
  }
}
