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.*;

public class B {
	public static void main(String[] args) throws IOException {
		Parser in = new Parser(System.in);
		int n = in.nextInt(), k = in.nextInt();
		int a[] = new int[n];
		int d[] = new int[n];
		a[0] = in.nextInt();
		for (int i = 1; i < n; i++) a[i] = in.nextInt();
		for (int i = 0; i < n; i++) {
			boolean f = false;
			for (int j = 0; j < n && !f; j++) {
				int tmp = a[i] + k*(j-i);
				f = f || (tmp <= 0);
				if (tmp > 0 && tmp == a[j]) d[i]++;
			}
			if (f) d[i] = 0;
		}
		int indmax = 0;
		for (int i = 1; i < n; i++)
			if (d[i] > d[indmax]) indmax = i;
		int c = 0;
		for (int i = 0; i < n; i++)
			if (a[indmax] + (i-indmax)*k != a[i]) c++;
		System.out.println(c);
		for (int i = 0; i < n; i++) {
			int tmp = a[indmax]+k*(i-indmax);
			if (a[i] != tmp)
				System.out.println(tmp > a[i] ? "+ " + (i+1) + " " + (tmp-a[i]) : "- " + (i+1) + " " + (a[i]-tmp));
		}
	}
}

class Parser {
	public Parser(InputStream in) {
		this.in = in;
	}

	public int nextInt() throws IOException {
		int c;
		while (((c = in.read()) < '0' || c > '9') && c != '-' && c != -1);
		int res = 0;;
		if (c != -1) {
			boolean neg = c == '-';
			if (neg) c = in.read();
			do res = res*10 + c - '0'; while ((c = in.read()) <= '9' && c >= '0');
			if (neg) res = -res;
		}
		return res;
	}

	private InputStream in;
}