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, a[1005], f[1005], las = 0;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  las = f[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (a[i] - (i - 1) * k <= 0) continue;
    f[i] = 1;
    for (int j = 1; j < i; ++j)
      if ((a[i] - a[j]) % k == 0 && (a[i] - a[j]) / k == (i - j))
        f[i] = max(f[j] + 1, f[i]);
    if (f[i] > f[las]) las = i;
  }
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    int t = a[las] + (i - las) * k;
    if (t != a[i]) ++cnt;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= n; ++i) {
    int t = a[las] + (i - las) * k;
    if (t < a[i])
      printf("- %d %d\n", i, a[i] - t);
    else if (t > a[i])
      printf("+ %d %d\n", i, t - a[i]);
  }
  return 0;
}
