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;
long long int a[1003], n, k, r[1003], mn = 999999999999, pos = 0, mo = 0,
                                      da = 0;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) r[i] = a[i] - (1 + i * k);
  for (int i = 0; i < n; i++) {
    mo = 0;
    for (int j = 0; j < n; j++)
      if (r[i] != r[j] && a[j] + (r[i] - r[j]) > 0)
        mo++;
      else if (a[j] + (r[i] - r[j]) <= 0)
        mo = 99999999999999;
    if (mo < mn) mn = mo, pos = i;
  }
  cout << mn << "\n";
  for (int i = 0; i < n; i++) {
    mo = r[pos] - r[i];
    if (mo == 0)
      continue;
    else {
      if (mo < 0)
        cout << "- ";
      else
        cout << "+ ";
      cout << i + 1 << " " << max(mo, mo * (-1)) << "\n";
    }
  }
  return 0;
}
