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() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  int ans = INT_MAX;
  int pos = 0;
  for (int i = 1; i <= 1000; i++) {
    int c = 0, p = 0;
    for (int j = i; c < n; j += k) {
      if (v[c++] != j) p++;
    }
    if (p < ans) pos = i, ans = p;
  }
  cout << ans << endl;
  if (ans > 0) {
    int c = 0;
    for (int i = pos; c < n; i += k) {
      if (v[c] != i) {
        int x = abs(i - v[c]);
        stringstream ss;
        string a, b;
        ss << c + 1;
        ss >> a;
        ss.clear();
        ss << x;
        ss >> b;
        ss.clear();
        string second;
        if (v[c] < i)
          second = "+ ";
        else
          second = "- ";
        cout << (second + a + " " + b) << endl;
      }
      c++;
    }
  }
  return 0;
}
