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 N = 1005;
int x[N];
int main() {
  int n, k;
  while (cin >> n >> k) {
    vector<pair<int, int> > ans, res;
    int max_x = 0;
    for (int i = 0; i < n; i++) {
      cin >> x[i];
      max_x = max(x[i], max_x);
    }
    int sum = 99999;
    for (int i = 1; i <= max_x; i++) {
      res.clear();
      int cont = 0;
      for (int j = 0; j < n; j++) {
        int temp = i + j * k - x[j];
        if (temp != 0) {
          res.push_back(make_pair(j + 1, temp));
          cont++;
        }
      }
      if (cont < sum) {
        sum = cont;
        ans.clear();
        ans.assign(res.begin(), res.end());
      }
    }
    cout << sum << endl;
    for (int i = 0; i < sum; i++) {
      if (ans[i].second > 0) {
        cout << "+ " << ans[i].first << " " << ans[i].second << endl;
      } else {
        int temp = abs(ans[i].second);
        cout << "- " << ans[i].first << ' ' << temp << endl;
      }
    }
  }
  return 0;
}
