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.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {public static void main(String[] args) throws Exception {new Solve();}}

class Solve { public Solve() throws Exception {solve();}
    static BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st = new StringTokenizer("");
    static Scanner sc = new Scanner(System.in);

    void solve() throws Exception {
        int n = NI();
        int k = NI();
        int a[] = new int[n];
        for (int i  =0; i < n; i++)
            a[i] = NI();

        int mres = Integer.MAX_VALUE, tres = 0, mind = 0;
       for (int i = 0; i < n; i++) {

           if (a[i] - i * k > 0) {
               tres = 0;
               for (int j = i - 1; j >= 0; j--) {
                   if (a[j] != a[i] - (i - j) * k)
                       tres++;
               }
               for (int j = i + 1; j < n; j++)
                   if (a[j] != a[i] + (j - i) * k)
                       tres++;
           }
           if (tres < mres) {
                mind = i;
               mres = tres;
           }
       }
        System.out.println(mres);
        for (int i = 0; i < mind; i++)
            if (a[i] != a[mind] - (mind - i) * k)
                System.out.println((a[i] > a[mind] - (mind - i) * k ? "- " : "+ ") + (i + 1) + " "+(abs(a[i] - (a[mind] - (mind - i) * k))));

        for (int i = mind + 1; i < n; i++)
            if (a[i] != a[mind] + (i - mind) * k)
                System.out.println((a[i] > a[mind] + (i - mind) * k ? "- " : "+ ") + (i + 1) +" " +(abs(a[i] - (a[mind] + (i - mind) * k))));
    }

    class Pair {
        int i, j;
        public Pair(int a, int b) {
            i = a;
            j = b;
        }
    }

    int min(int i1,int i2){return i1<i2?i1:i2;}
    long min(long i1,long i2){return i1<i2?i1:i2;}
    int max(int i1,int i2){return i1>i2?i1:i2;}
    long max(long i1,long i2){return i1>i2?i1:i2;}
    String NS()throws Exception{while(!st.hasMoreTokens())st=new StringTokenizer(stdin.readLine());return st.nextToken();}
    String NLn() throws Exception {return stdin.readLine();}
    int NI()throws Exception{return Integer.parseInt(NS());}
    long NL()throws Exception{return Long.parseLong(NS());}
    double ND()throws Exception{return Double.parseDouble(NS());}
    int abs(int x){return x<0?-x:x;}
    void sortA(int[]a){Arrays.sort(a);}
}