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;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int n, k;
  cin >> n >> k;
  int arr[n + 1];
  for (int i = 1; i <= n; i++) cin >> arr[i];
  vector<string> ans(n + 100, "");
  for (int i = 1; i <= 1000; i++) {
    vector<string> cur;
    for (int j = 1; j <= n; j++) {
      int val = i + (j - 1) * k;
      string s = "";
      if (val < arr[j]) {
        s = "- ";
        s += to_string(j);
        s += ' ';
        s += to_string(arr[j] - val);
        cur.push_back(s);
      } else if (val > arr[j]) {
        s = "+ ";
        s += to_string(j);
        s += ' ';
        s += to_string(val - arr[j]);
        cur.push_back(s);
      }
    }
    if (cur.size() < ans.size()) ans = cur;
  }
  cout << ans.size() << '\n';
  for (string i : ans) cout << i << '\n';
}
