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 n, k, ans, inp, res;
vector<int> s;
void check(int x) {
  int temp = 0;
  for (int i = 0; i < n; i++) {
    if ((s[i] - s[x]) != k * (i - x)) {
      temp++;
    }
  }
  if (temp < ans) {
    ans = temp;
    res = x;
  }
}
bool valid(int x) {
  if (s[x] - x * k <= 0) return false;
  return true;
}
void solve() {
  int temp = 0;
  for (int i = 0; i < n; i++) {
    if ((s[i] - s[res]) != k * (i - res)) {
      if (s[i] < s[res] + k * (i - res))
        cout << "+ " << i + 1 << " " << s[res] + k * (i - res) - s[i] << "\n";
      else
        cout << "- " << i + 1 << " " << s[i] - (s[res] + k * (i - res)) << "\n";
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> inp;
    s.push_back(inp);
  }
  ans = n - 1;
  res = 0;
  for (int i = 1; i < n; i++) {
    if (valid(i)) check(i);
  }
  cout << ans << "\n";
  ;
  solve();
}
