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;
struct debug_t {
  template <typename T>
  debug_t& operator<<(const T& value) {
    return *this;
  }
} debug;
unsigned short n, k;
int a[1000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  unsigned short i, j;
  for (i = 0; i < n; i++) cin >> a[i];
  unsigned short result = USHRT_MAX, sub, best;
  for (i = 0; i < n; i++) {
    int need = a[i];
    sub = 0;
    j = i;
    while (j--) {
      need -= k;
      if (!(need <= 0)) {
      } else
        break;
      if (a[j] != need) sub++;
    }
    if (need > 0) {
      j = i;
      need = a[i];
      while (++j < n) {
        need += k;
        if (a[j] != need) sub++;
      }
      if (sub < result) {
        result = sub;
        best = i;
      }
    }
  }
  cout << result << '\n';
  i = best;
  int need = a[i];
  while (i--) {
    need -= k;
    if (a[i] > need)
      cout << "- " << i + 1 << ' ' << a[i] - need << '\n';
    else if (a[i] < need)
      cout << "+ " << i + 1 << ' ' << need - a[i] << '\n';
  }
  i = best;
  need = a[i];
  while (++i < n) {
    need += k;
    if (a[i] > need)
      cout << "- " << i + 1 << ' ' << a[i] - need << '\n';
    else if (a[i] < need)
      cout << "+ " << i + 1 << ' ' << need - a[i] << '\n';
  }
}
