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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> V(n);
  for (int i = 0; i < n; ++i) cin >> V[i];
  vector<int> A, B;
  int ans = 1e9;
  for (int i = 1; i <= 1000; i++) {
    int cur = i, cnt = 0;
    vector<int> auxA, auxB;
    for (int j = 0; j < n; ++j) {
      if (cur != V[j]) {
        auxA.push_back(j);
        auxB.push_back(cur);
        cnt++;
      }
      cur += k;
    }
    if (ans > cnt) {
      ans = cnt;
      A = auxA, B = auxB;
    }
  }
  cout << ans << "\n";
  for (int i = 0; i < ans; ++i) {
    if (V[A[i]] > B[i])
      cout << "- " << A[i] + 1 << " " << V[A[i]] - B[i] << "\n";
    else
      cout << "+ " << A[i] + 1 << " " << B[i] - V[A[i]] << "\n";
  }
  return 0;
}
