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 N = 1e3 + 5;
int n, k, a[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int ans = n, st;
  for (int i = 1; i <= 1000; ++i) {
    int sum = 0;
    for (int x = i, j = 1; j <= n; ++j, x += k) sum += (a[j] != x);
    if (sum < ans) ans = sum, st = i;
  }
  printf("%d\n", ans);
  for (int x = st, i = 1; i <= n; ++i, x += k)
    if (x != a[i]) {
      printf("%c %d %d\n", x < a[i] ? '-' : '+', i, std::abs(a[i] - x));
    }
  return 0;
}
