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 main() {
  long long n, k;
  scanf("%lld %lld", &n, &k);
  long long arr[n + 1];
  for (int i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
  }
  long long ans = LONG_LONG_MAX, pos;
  for (int i = 0; i < n; i++) {
    long long cnt = 0;
    if (arr[i] - i * k <= 0) continue;
    for (int j = i + 1; j < n; j++) {
      if (arr[j] - arr[i] != (j - i) * k) cnt++;
    }
    for (int j = i - 1; j >= 0; j--) {
      if (arr[i] - arr[j] != (i - j) * k) {
        cnt++;
      }
    }
    if (ans > cnt) {
      ans = cnt;
      pos = i;
    }
  }
  cout << ans << endl;
  for (long long i = 0; i < n; i++) {
    if (i < pos) {
      if (arr[pos] - arr[i] > ((pos - i) * k)) {
        printf("+ %lld %lld\n", i + 1, arr[pos] - arr[i] - ((pos - i) * k));
      } else if (arr[pos] - arr[i] < ((pos - i) * k)) {
        printf("- %lld %lld\n", i + 1, ((pos - i) * k) - arr[pos] + arr[i]);
      }
    }
    if (i > pos) {
      if (arr[i] - arr[pos] > ((i - pos) * k)) {
        printf("- %lld %lld\n", i + 1, arr[i] - arr[pos] - ((i - pos) * k));
      } else if (arr[i] - arr[pos] < ((i - pos) * k)) {
        printf("+ %lld %lld\n", i + 1, ((i - pos) * k) - arr[i] + arr[pos]);
      }
    }
  }
  return 0;
}
