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 n, k, a[1010];
vector<pair<char, pair<int, int> > > ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  ans.push_back({'#', {0, 0}});
  for (int i = 1; i <= n; i++) cin >> a[i], ans.push_back({'#', {0, 0}});
  for (int x = 1; x <= 3000; x++) {
    vector<pair<char, pair<int, int> > > tmp;
    if (a[1] != x) {
      char ch = '+';
      if (a[1] > x) ch = '-';
      tmp.push_back({ch, {1, abs(a[1] - x)}});
    }
    for (int i = 2; i <= n; i++) {
      char ch = '+';
      int y = x + (i - 1) * k;
      if (a[i] == y) continue;
      if (a[i] > y) ch = '-';
      tmp.push_back({ch, {i, abs(a[i] - y)}});
    }
    if (tmp.size() < ans.size()) ans = tmp;
  }
  cout << ans.size() << endl;
  for (pair<char, pair<int, int> > x : ans)
    cout << x.first << " " << x.second.first << " " << x.second.second << endl;
  return 0;
}
