#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
int aa[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> aa[i];
  int ans = N, ind = -1;
  for (int i = 1; i <= n; i++) {
    int curAns = 0;
    for (int j = i - 1; j >= 1; j--) {
      int should = aa[i] - (i - j) * k;
      if (should <= 0) curAns += N;
      if (should != aa[j]) curAns++;
    }
    for (int j = i + 1; j <= n; j++) {
      int should = aa[i] + (j - i) * k;
      if (should != aa[j]) curAns++;
    }
    if (curAns < ans) {
      ans = curAns;
      ind = i;
    }
  }
  assert(ans != N);
  cout << ans << "\n";
  int curAns = 0;
  for (int j = ind - 1; j >= 1; j--) {
    int should = aa[ind] - (ind - j) * k;
    if (should > aa[j]) cout << "+ " << j << " " << should - aa[j] << "\n";
    if (should < aa[j]) cout << "- " << j << " " << aa[j] - should << "\n";
  }
  for (int j = ind + 1; j <= n; j++) {
    int should = aa[ind] + (j - ind) * k;
    if (should > aa[j]) cout << "+ " << j << " " << should - aa[j] << "\n";
    if (should < aa[j]) cout << "- " << j << " " << aa[j] - should << "\n";
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
int aa[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> aa[i];
  int ans = N, ind = -1;
  for (int i = 1; i <= n; i++) {
    int curAns = 0;
    for (int j = i - 1; j >= 1; j--) {
      int should = aa[i] - (i - j) * k;
      if (should <= 0) curAns += N;
      if (should != aa[j]) curAns++;
    }
    for (int j = i + 1; j <= n; j++) {
      int should = aa[i] + (j - i) * k;
      if (should != aa[j]) curAns++;
    }
    if (curAns < ans) {
      ans = curAns;
      ind = i;
    }
  }
  assert(ans != N);
  cout << ans << "\n";
  int curAns = 0;
  for (int j = ind - 1; j >= 1; j--) {
    int should = aa[ind] - (ind - j) * k;
    if (should > aa[j]) cout << "+ " << j << " " << should - aa[j] << "\n";
    if (should < aa[j]) cout << "- " << j << " " << aa[j] - should << "\n";
  }
  for (int j = ind + 1; j <= n; j++) {
    int should = aa[ind] + (j - ind) * k;
    if (should > aa[j]) cout << "+ " << j << " " << should - aa[j] << "\n";
    if (should < aa[j]) cout << "- " << j << " " << aa[j] - should << "\n";
  }
  return 0;
}

```