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 n;
int k;
int arr[2000];
int back(int a) {
  int c = 0;
  for (int i = a - 1; i >= 0; i--) {
    if (arr[a] - arr[i] != k * (a - i)) c++;
  }
  return c;
}
int front(int a) {
  int c = 0;
  for (int i = a + 1; i < n; i++) {
    if (arr[i] - arr[a] != k * (i - a)) c++;
  }
  return c;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> arr[i];
  int p, mi = 999999;
  for (int i = 0; i < n; i++) {
    if (front(i) + back(i) < mi && arr[i] > k * i) {
      p = i;
      mi = front(i) + back(i);
    }
  }
  cout << mi << endl;
  for (int i = p - 1; i >= 0; i--) {
    if (arr[p] - arr[i] != k * (p - i)) {
      if (arr[p] - arr[i] > k * (p - i)) {
        cout << "+"
             << " " << i + 1 << " " << abs(arr[p] - arr[i] - k * (p - i));
      } else
        cout << "-"
             << " " << i + 1 << " "
             << "" << abs(arr[p] - arr[i] - k * (p - i));
      cout << endl;
    }
  }
  for (int i = p + 1; i < n; i++) {
    if (arr[i] - arr[p] != k * (i - p)) {
      if (arr[i] - arr[p] > k * (i - p))
        cout << "-"
             << " " << i + 1 << " "
             << "" << abs(arr[p] - arr[i] - k * (p - i));
      else
        cout << "+"
             << " " << i + 1 << " " << abs(arr[p] - arr[i] - k * (p - i));
      cout << endl;
    }
  }
  return 0;
}
