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 arr[1010], prr[1010], brr[1010];
int main() {
  ios_base::sync_with_stdio(0);
  long long n, k, max_h, i, b, j, a;
  cin >> n >> k;
  max_h = -1;
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
    max_h = max(max_h, arr[i]);
  }
  for (i = 1; i <= max_h; i++) {
    fill(brr + 1, brr + 1 + n, 0);
    if (arr[1] > i)
      brr[1] -= arr[1] - i;
    else
      brr[1] = i - arr[1];
    b = (brr[1] != 0);
    for (j = 2; j <= n; j++) {
      if (arr[j] > arr[j - 1] + brr[j - 1] + k)
        brr[j] -= arr[j] - (arr[j - 1] + brr[j - 1] + k);
      else
        brr[j] = (arr[j - 1] + brr[j - 1] + k) - arr[j];
      b += (brr[j] != 0);
    }
    if ((i == 1) || (a > b)) {
      a = b;
      for (j = 1; j <= n; j++) {
        prr[j] = brr[j];
      }
    }
  }
  cout << a << '\n';
  for (i = 1; i <= n; i++) {
    if (prr[i] > 0) cout << "+ " << i << " " << prr[i] << '\n';
    if (prr[i] < 0) cout << "- " << i << " " << -prr[i] << '\n';
  }
  return 0;
}
