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.io.IOException;
import java.util.Arrays;
import java.util.Scanner;

/**
 * TODO: describe
 *
 * @author keks
 */
public class B {
    public static void main(String[] args) throws IOException {
        final Scanner sc = new Scanner(System.in);
        final int n = sc.nextInt();
        final int k = sc.nextInt();
        final int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        final int[] res = get(n, k, a);
        final int diff = res[0];
        final int count = res[1];
        System.out.println(count);
        for (int i = 0; i < n; i++) {
            if (a[i] != k * i + diff) {
                final int d = k * i + diff - a[i];
                System.out.println((d > 0 ? '+' : '-') + " " + (i + 1) + " " + Math.abs(d));
            }
        }
    }

    public static int[] get(int n, int k, int[] a) {
        final int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            b[i] = i * k;
        }

        for (int i = 0; i < n; i++) {
            b[i] = a[i] - b[i];
        }

        Arrays.sort(b);

        int max = 0;
        int maxCount = 0;

        int count = 0;
        int current = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            if (b[i] != current) {
                current = b[i];
                count = 1;
            } else {
                count++;
            }

            if (count > maxCount && (current == Integer.MIN_VALUE || current > 0)) {
                maxCount = count;
                max = current;
            }
        }

        return new int[]{max, n - maxCount};
    }
}
