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
import java.util.Scanner;


public class B402 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), k = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = in.nextInt();
        int bestTime = Integer.MAX_VALUE, bestFirst = 0;
        for (int first = 1; first <= 1000; first++) {
            int currentTime = 0;
            for (int i = 0; i < n; i++)
                if (first + i * k != a[i])
                    currentTime++;
            if (currentTime < bestTime) {
                bestTime = currentTime;
                bestFirst = first;
            }
        }
        System.out.println(bestTime);
        for (int i = 0; i < n; i++) {
            int need = bestFirst + i * k;
            if (need != a[i])
                System.out.println((need > a[i] ? "+ " : "- ") + (i + 1) + " " + Math.abs(need - a[i]));
        }
    }
}
