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 dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  int a[1001];
  for (long long i = 0; i < n; i++) cin >> a[i];
  int res = n;
  int id = 0;
  for (long long i = 0; i < n; i++) {
    int x = 0;
    if (k * i >= a[i]) continue;
    for (long long j = 0; j < n; j++) {
      if (i == j) continue;
      if (j < i) {
        if (a[j] != a[i] - (i - j) * k) x++;
      } else {
        if (a[j] != a[i] + (j - i) * k) x++;
      }
    }
    if (x < res) {
      res = x;
      id = i;
    }
  }
  cout << res << endl;
  for (long long i = 0; i < n; i++) {
    if (i == id) continue;
    if (i < id) {
      if (a[i] != a[id] - (id - i) * k) {
        if (a[i] > a[id] - (id - i) * k)
          cout << "-";
        else
          cout << "+";
        cout << " " << i + 1 << " " << abs(a[i] - (a[id] - (id - i) * k))
             << endl;
      }
    } else {
      if (a[i] != a[id] + (i - id) * k) {
        if (a[i] < a[id] + (i - id) * k)
          cout << "+ ";
        else
          cout << "- ";
        cout << i + 1 << " " << abs(a[i] - (a[id] + (i - id) * k)) << endl;
      }
    }
  }
  return 0;
}
