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>
#pragma warning(disable : 4996)
using namespace std;
int a[2005];
int b[2005];
int main() {
  int n, k;
  while (cin >> n >> k) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int mmin = n + 5, wh = 0;
    int num = 0;
    for (int i = 1; i <= n; i++) {
      num = 0;
      if (a[i] - (i - 1) * k < 1) continue;
      b[i] = a[i];
      for (int j = i - 1; j >= 1; j--) {
        b[j] = b[j + 1] - k;
        if (b[j] != a[j]) num++;
      }
      for (int j = i + 1; j <= n; j++) {
        b[j] = b[j - 1] + k;
        if (b[j] != a[j]) num++;
      }
      if (num < mmin) {
        mmin = num;
        wh = i;
      }
    }
    printf("%d\n", mmin);
    b[wh] = a[wh];
    for (int j = wh - 1; j >= 1; j--) {
      b[j] = b[j + 1] - k;
    }
    for (int j = wh + 1; j <= n; j++) {
      b[j] = b[j - 1] + k;
    }
    for (int i = 1; i <= n; i++) {
      if (b[i] > a[i])
        printf("+ %d %d\n", i, b[i] - a[i]);
      else if (b[i] < a[i])
        printf("- %d %d\n", i, -b[i] + a[i]);
    }
  }
  return 0;
}
