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.*;
import java.util.*;
public class b
{
	public static void main(String[]args)
	{
		Scanner N=new Scanner(System.in);
		int n=N.nextInt();
		int k=N.nextInt();
		int ar[]=new int[n];
		int ac[]=new int[n];
		for(int i=0;i<n;i++)ar[i]=N.nextInt();
		Arrays.fill(ac,1);
		int pos;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<i;j++)
			{
				if(ar[i]-ar[j]==k*(i-j))
				ac[i]++;
			}
			for(int j=i+1;j<n;j++)
			{
				if(ar[j]-ar[i]==k*(j-i))
				ac[i]++;
			}
		}
		int mx=ac[0];
		pos=0;
		for(int i=0;i<n;i++)
		{
			if(ac[i]>mx)
			{
				if(ar[i]-k*(i)>0)
				{
					mx=ac[i];
					pos=i;
				}
			}
		}
		if(ac[pos]==n)
		{System.out.println("0");System.exit(0);}
		int ans[]=new int[n];
		ans[0]=ar[pos]-k*pos;
		for(int i=1;i<n;i++)ans[i]=ans[i-1]+k;
		System.out.println(n-mx);
		for(int i=0;i<n;i++)
		{
			
			if(ar[i]==ans[i])
			continue;
			if(ar[i]>ans[i])
			System.out.println("- "+(i+1)+" "+Math.abs(ans[i]-ar[i]));
			else
			System.out.println("+ "+(i+1)+" "+Math.abs(ans[i]-ar[i]));

		}
	}
}