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;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> v;
  int ans = 2000;
  for (int i = 1; i <= 1000; i++) {
    vector<int> b(n);
    b[0] = i;
    for (int j = 1; j < n; j++) {
      b[j] = b[j - 1] + k;
    }
    int cur = 0;
    for (int j = 0; j < n; j++) {
      cur += (a[j] != b[j]);
    }
    if (cur < ans) {
      ans = cur;
      v = b;
    }
  }
  cout << ans << "\n";
  for (int i = 0; i < n; i++) {
    if (a[i] < v[i]) {
      cout << "+ " << i + 1 << " " << v[i] - a[i] << "\n";
    } else if (a[i] > v[i]) {
      cout << "- " << i + 1 << " " << a[i] - v[i] << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}
