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;
int h[1001];
int p[1001];
int ans, pos;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &h[i]);
  ans = 999999;
  for (int i = 1; i <= n; i++) {
    int num = 0;
    p[i] = h[i];
    int l = i - 1;
    int r = i + 1;
    int key = 0;
    while (l > 0) {
      if (h[l] != p[l + 1] - k) num++;
      p[l] = p[l + 1] - k;
      if (p[l] <= 0) {
        key = 1;
        break;
      }
      l--;
    }
    if (key == 1) continue;
    while (r <= n) {
      if (h[r] != p[r - 1] + k) num++;
      p[r] = p[r - 1] + k;
      r++;
    }
    if (num < ans) {
      ans = num;
      pos = i;
    }
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    int high = h[pos] + (i - pos) * k;
    if (h[i] > high) {
      cout << "- " << i << " " << h[i] - high << endl;
    }
    if (h[i] < high) {
      cout << "+ " << i << " " << high - h[i] << endl;
    }
  }
  return 0;
}
