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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class TreesRow {
    void solve() {
        int n = in.nextInt(), k = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = in.nextInt();
        
        int ans = n + 1;
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            if (a[i] - i * k <= 0) continue;
            
            int cnt = 0;
            int[] d = new int[n];
            
            int p = i, q = i, cur = a[i];
            while (p >= 0) {
                d[p] = cur - a[p];
                if (d[p] != 0) cnt++;
                cur -= k;
                p--;
            }
            cur = a[i];
            while (q < n) {
                d[q] = cur - a[q];
                if (d[q] != 0) cnt++;
                cur += k;
                q++;
            }
            
            if (cnt < ans) {
                ans = cnt;
                res = d;
            }
        }
        
        out.println(ans);
        for (int i = 0; i < n; i++) {
            if (res[i] > 0) {
                out.printf("+ %d %d%n", i + 1, res[i]);
            } else if (res[i] < 0) {
                out.printf("- %d %d%n", i + 1, -res[i]);
            }
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new TreesRow().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
