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 n, k, pos, res, cur;
int a[1005];
int main() {
  res = 1000000;
  scanf("%d %d\n", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d ", &a[i]);
  for (int i = 0; i < n; i++) {
    int res1 = 0, cur = a[i];
    bool ok = true;
    for (int j = i - 1; j >= 0 && ok; j--) {
      if (a[j] != cur - k && cur - k > 0)
        res1++;
      else
        ok = false;
      cur -= k;
    }
    if (!ok) continue;
    cur = a[i];
    for (int j = i + 1; j < n; j++) {
      if (a[j] != cur + k) res1++;
      cur += k;
    }
    if (res > res1) res = res1, pos = i;
  }
  cur = a[pos];
  printf("%d\n", res);
  for (int i = pos - 1; i >= 0; i--) {
    cur -= k;
    int x = cur - a[i];
    if (x != 0) printf("%c %d %d\n", (x > 0 ? '+' : '-'), i + 1, abs(x));
  }
  cur = a[pos];
  for (int i = pos + 1; i < n; i++) {
    cur += k;
    int x = cur - a[i];
    if (x != 0) printf("%c %d %d\n", (x > 0 ? '+' : '-'), i + 1, abs(x));
  }
  return 0;
}
