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;
struct arr {
  char c;
  int u;
  int v;
};
struct arr q[1005];
int main() {
  int n, k, i, m = 0, j, t, l = 99999999, p = 0, s = 0;
  int a[1005];
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    m = max(m, a[i] + (k * (n - i - 1)));
  }
  for (i = m; i >= (1 + (n - 1) * k); i--) {
    s = 0;
    t = i;
    for (j = n - 1; j >= 0; j--) {
      if (abs(a[j] - t) != 0) s++;
      t = t - k;
    }
    if (s < l) {
      l = s;
      p = i;
    }
  }
  t = 1;
  l = 0;
  for (i = n - 1; i >= 0; i--) {
    if (a[i] < (p)) {
      q[l].c = '+';
      q[l].u = i + 1;
      q[l].v = abs(a[i] - p);
      l++;
    } else if (a[i] > (p)) {
      q[l].c = '-';
      q[l].u = i + 1;
      q[l].v = abs(a[i] - p);
      l++;
    }
    p = p - k;
  }
  cout << l << endl;
  for (i = l - 1; i >= 0; i--) {
    cout << q[i].c << " " << q[i].u << " " << q[i].v << endl;
  }
  return 0;
}
