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 ar[1001];
bool veri(vector<int> x) {
  for (int i = 0; i < x.size(); i++) {
    if (x[i] <= 0) {
      return false;
    }
  }
  return true;
}
int distintos(vector<int> x) {
  int cnt = 0;
  for (int i = 0; i < x.size(); i++) {
    if (x[i] != ar[i]) {
      cnt++;
    }
  }
  return cnt;
}
int main() {
  int n, y;
  cin >> n >> y;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  int mini = n + 1;
  vector<int> save;
  for (int i = 0; i < n; i++) {
    vector<int> x(n, 0);
    int var = x[i];
    x[i] = ar[i];
    for (int k = i - 1; k >= 0; k--) {
      x[k] = x[k + 1] - y;
    }
    for (int k = i + 1; k < n; k++) {
      x[k] = x[k - 1] + y;
    }
    if (veri(x)) {
      int dis = distintos(x);
      if (dis < mini) {
        mini = dis;
        save = x;
      }
    }
  }
  cout << mini << endl;
  for (int i = 0; i < save.size(); i++) {
    if (save[i] != ar[i]) {
      if (save[i] > ar[i]) {
        cout << "+ " << i + 1 << " " << save[i] - ar[i] << endl;
      } else {
        cout << "- " << i + 1 << " " << ar[i] - save[i] << endl;
      }
    }
  }
}
