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;
vector<pair<char, pair<int, int> > > v;
int main() {
  int n, k;
  int ans = 100000000;
  vector<int> vec;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    int temp1;
    scanf("%d", &temp1);
    vec.push_back(temp1);
  }
  for (int i = 1; i < 1001; i++) {
    int temp = 0;
    vector<pair<char, pair<int, int> > > tem;
    for (int j = 0; j < n; j++) {
      int a;
      a = abs(vec[j] - (i + j * k));
      if (a != 0) {
        temp++;
        if (vec[j] > i + j * k) {
          tem.push_back(pair<char, pair<int, int> >(
              '-', pair<int, int>(j + 1, vec[j] - (i + j * k))));
        } else {
          tem.push_back(pair<char, pair<int, int> >(
              '+', pair<int, int>(j + 1, -vec[j] + i + j * k)));
        }
      }
    }
    if (temp < ans) {
      v.clear();
      ans = temp;
      for (int j = 0; j < tem.size(); j++) {
        v.push_back(tem[j]);
      }
    }
  }
  cout << ans << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%c %d %d\n", v[i].first, v[i].second.first, v[i].second.second);
  }
}
