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 i, j, t;
  vector<int> a;
  for (i = 0; i < n; i++) {
    cin >> t;
    a.push_back(t);
  }
  int index = 0, s = 0, ans = 1e9;
  for (i = 0; i < n; i++) {
    int v = a[i], d = a[i];
    s = 0;
    bool f = true;
    for (j = i - 1; j >= 0; j--) {
      v -= k;
      if (v <= 0) {
        f = false;
        break;
      }
      if (v == a[j]) continue;
      s++;
    }
    if (!f) continue;
    for (j = i + 1; j < n; j++) {
      d += k;
      if (d == a[j]) continue;
      s++;
    }
    if (s < ans) {
      index = i;
      ans = s;
    }
  }
  cout << ans << endl;
  int v = a[index], d = a[index];
  for (j = index - 1; j >= 0; j--) {
    v -= k;
    if (v == a[j]) continue;
    if (a[j] < v)
      cout << '+' << " " << j + 1 << " " << v - a[j] << endl;
    else
      cout << '-' << " " << j + 1 << " " << a[j] - v << endl;
  }
  for (j = index + 1; j < n; j++) {
    d += k;
    if (d == a[j]) continue;
    if (a[j] < d)
      cout << '+' << " " << j + 1 << " " << d - a[j] << endl;
    else
      cout << '-' << " " << j + 1 << " " << a[j] - d << endl;
  }
  return 0;
}
