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
#include <bits/stdc++.h>
using namespace std;
int maxi(int x, int y) { return (x > y) ? x : y; }
int mini(int x, int y) { return (x > y) ? y : x; }
vector<vector<int> > a;
vector<int> q;
int arr[10000];
int main() {
  int n, k, temp;
  scanf("%d %d", &n, &k);
  ;
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  ;
  for (int i = 1; i <= 1000; i++) {
    temp = i;
    q.clear();
    for (int j = 1; j <= n; j++) {
      q.push_back(temp);
      temp = temp + k;
    }
    a.push_back(q);
  }
  int temp3 = INT_MIN;
  int temp2, cnt;
  for (int i = 0; i < a.size(); i++) {
    vector<int> b = a[i];
    cnt = 0;
    for (int j = 0; j < b.size(); j++) {
      if (arr[j] == b[j]) cnt++;
    }
    if (cnt > temp3) {
      temp3 = cnt;
      temp2 = i;
    }
  }
  vector<int> p = a[temp2];
  if (temp3 == n) {
    printf("0\n");
    return 0;
  }
  printf("%d\n", n - temp3);
  for (int i = 0; i < p.size(); i++) {
    if (arr[i] < p[i])
      printf("+ %d %d\n", i + 1, p[i] - arr[i]);
    else if (arr[i] > p[i])
      printf("- %d %d\n", i + 1, arr[i] - p[i]);
    else
      ;
  }
  return 0;
}
