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 b[1005];
int a[1005];
int slove(int *b, int n, int k) {
  for (int i = 0; i <= n; ++i) a[i] = b[i];
  int res = 0;
  for (int i = 1; i <= n; ++i) {
    if ((a[i] - a[i - 1]) != k) {
      res++;
      a[i] = a[i - 1] + k;
    }
  }
  return res;
}
int slove2(int *b, int n, int k) {
  for (int i = 0; i <= n; ++i) a[i] = b[i];
  int res = 0;
  for (int i = 1; i <= n; ++i) {
    if ((a[i] - a[i - 1]) != k) {
      if (a[i] > a[i - 1] + k)
        printf("- %d %d\n", i, a[i] - k - a[i - 1]);
      else
        printf("+ %d %d\n", i, a[i - 1] + k - a[i]);
      a[i] = a[i - 1] + k;
      res++;
    }
  }
  return res;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> b[i];
  int best = INT_MAX;
  int no = 0;
  for (int i = -k + 1; i <= 1005; i++) {
    b[0] = i;
    int cur = slove(b, n, k);
    if (best >= cur) {
      best = cur;
      no = i;
    }
  }
  cout << best << endl;
  b[0] = no;
  slove2(b, n, k);
  return 0;
}
