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;
const int INF = 1e9 + 9;
int n, k;
vector<int> a;
int f(int st) {
  int r = 0;
  for (int i = 0; i < n; ++i)
    if (a[i] != st + i * k) ++r;
  return r;
}
int main() {
  cin >> n >> k;
  a.resize(n);
  int r = -1;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (r < a[i]) r = a[i];
  }
  int res = n + 5, wh;
  for (int i = 1; i <= r; ++i) {
    int t = f(i);
    if (t < res) {
      res = t;
      wh = i;
    }
  }
  cout << res << endl;
  for (int i = 0; i < n; ++i)
    if (a[i] > wh + i * k)
      cout << "- " << i + 1 << ' ' << a[i] - wh - i * k << endl;
    else if (a[i] < wh + i * k)
      cout << "+ " << i + 1 << ' ' << -a[i] + wh + i * k << endl;
  cout << "";
  return 0;
}
