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 i, j, k, u, v, m, n, ma, ans = 9999999, ans1;
int a[5001] = {};
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    ma = max(ma, a[i] + k);
  }
  for (i = 1; i <= ma; i++) {
    int s = 0;
    int m = i;
    for (j = 1; j <= n; j++) {
      if (a[j] != m) s++;
      m += k;
    }
    if (m < 0) continue;
    if (ans > s) {
      ans = s;
      ans1 = i;
    }
  }
  cout << ans << endl;
  if (ans)
    for (i = 1; i <= n; i++) {
      if (ans1 != a[i]) {
        if (a[i] < ans1)
          cout << "+" << i << " " << ans1 - a[i] << endl;
        else
          cout << "-" << i << " " << a[i] - ans1 << endl;
      }
      ans1 += k;
    }
  return 0;
}
