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>
const int MAXN = 1005;
int a[MAXN];
int cnt[MAXN];
int main() {
  int n, k;
  while (scanf("%d%d", &n, &k) + 1) {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if ((j - i) * k == a[j] - a[i]) {
          cnt[i]++;
        }
      }
    }
    int mm = 0;
    int p = 0;
    for (int i = 1; i <= n; i++) {
      if (cnt[i] > mm && a[i] - (i - 1) * k > 0) {
        mm = cnt[i];
        p = i;
      }
    }
    printf("%d\n", n - mm);
    for (int i = 1; i <= n; i++) {
      if ((i - p) * k != a[i] - a[p]) {
        if (a[i] - a[p] > (i - p) * k) {
          printf("- %d %d\n", i, a[i] - a[p] - (i - p) * k);
        } else {
          printf("+ %d %d\n", i, (i - p) * k - a[i] + a[p]);
        }
      }
    }
  }
  return 0;
}
