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;
int main() {
  long long int n, k, t, s, x, i, count;
  vector<long long int> a;
  vector<pair<long long int, pair<long long int, long long int> > > r;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  long long int mcount = -1;
  long long int ind = 0;
  for (long long int i = 0; i < n; i++) {
    count = 1;
    t = 1;
    for (long long int j = i + 1; j < n; j++) {
      if (a[j] == a[i] + t * k) count++;
      t++;
    }
    if (count > mcount && a[i] > i * k) {
      mcount = count;
      ind = i;
    }
  }
  vector<long long int> res(n);
  res[ind] = a[ind];
  for (long long int i = ind - 1; i >= 0; i--) {
    res[i] = res[i + 1] - k;
  }
  for (long long int i = ind + 1; i < n; i++) {
    res[i] = res[i - 1] + k;
  }
  for (long long int i = 0; i < n; i++) {
    if (a[i] > res[i]) {
      s = a[i] - res[i];
      r.push_back(make_pair(-1, make_pair(i + 1, s)));
    } else if (a[i] < res[i]) {
      s = res[i] - a[i];
      r.push_back(make_pair(1, make_pair(i + 1, s)));
    }
  }
  cout << r.size() << endl;
  for (int i = 0; i < (int)r.size(); i++) {
    if (r[i].first == 1)
      cout << "+ ";
    else
      cout << "- ";
    cout << r[i].second.first << " " << r[i].second.second << endl;
  }
  return 0;
}
