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() {
  int i, j, k, n, t, sig, tsig, mx;
  int array[1001];
  while (cin >> n >> k) {
    memset(array, 0, sizeof(array));
    for (i = 0; i < n; i++) {
      scanf("%d", &t);
      if (i == 0)
        mx = t;
      else
        mx = mx > t ? mx : t;
      array[i] = t;
    }
    sig = 0, t = 1;
    for (i = 1; i <= mx; i++) {
      for (tsig = j = 0; j < n; j++) {
        if (array[j] == i + j * k) tsig++;
      }
      if (sig < tsig) t = i, sig = tsig;
    }
    for (j = i = 0; i < n; i++) {
      if (array[i] != t + i * k) j++;
    }
    cout << j << endl;
    for (i = 0; i < n; i++) {
      if (array[i] != t + i * k)
        mx = t + i * k - array[i];
      else
        continue;
      if (mx > 0)
        printf("+ %d %d\n", i + 1, mx);
      else
        printf("- %d %d\n", i + 1, -mx);
    }
  }
  return 0;
}
