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 calc[1001][1001];
int c[1001];
int arr[1001];
int main() {
  int n, k;
  cin >> n >> k;
  int i;
  memset(c, 0, sizeof(c));
  memset(calc, 0, sizeof(calc));
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  int num = 0;
  int j;
  for (i = 1; i <= n; i++) {
    num = arr[i];
    if (arr[i] - ((i - 1) * k) <= 0) {
      c[i] = 99999999;
      continue;
    }
    for (j = i - 1; j >= 1; j--) {
      num -= k;
      if (arr[j] != num) {
        c[i]++;
        calc[i][j] = num - arr[j];
      }
    }
    num = arr[i];
    for (j = i + 1; j <= n; j++) {
      num += k;
      if (arr[j] != num) {
        c[i]++;
        calc[i][j] = num - arr[j];
      }
    }
  }
  int min1 = c[1];
  int pos = 1;
  for (i = 1; i <= n; i++) {
    if (c[i] < min1) {
      min1 = c[i];
      pos = i;
    }
  }
  cout << min1 << "\n";
  for (i = 1; i <= n; i++) {
    if (calc[pos][i] > 0)
      cout << "+"
           << " " << i << " " << calc[pos][i] << "\n";
    else if (calc[pos][i] < 0) {
      cout << "-"
           << " " << i << " " << abs(calc[pos][i]) << "\n";
    }
  }
  return 0;
}
