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 MAXH = 1010;
int n, k;
vector<int> arr;
vector<int> ans;
int best = -1, bestS0 = 0;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    arr.push_back(x);
  }
  for (int s0 = 1; s0 < MAXH; s0++) {
    int res = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] == s0 + k * i) {
        res++;
      }
    }
    if (res > best) {
      best = res;
      bestS0 = s0;
    }
  }
  cout << (n - best) << endl;
  for (int i = 0; i < n; i++) {
    int d = (bestS0 + i * k) - arr[i];
    if (d != 0) {
      if (d > 0)
        cout << "+ ";
      else
        cout << "- ";
      cout << (i + 1) << " ";
      cout << abs(d) << endl;
    }
  }
  return 0;
}
