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;
int a[1100];
int main() {
  cin >> n >> K;
  int i, j, k;
  int res = 1000000, ans;
  for (i = 1; i <= n; ++i) cin >> a[i];
  for (i = 1; i <= 1000; ++i) {
    int c = 0;
    for (j = 1, k = i; j <= n; ++j, k += K)
      if (a[j] != k) ++c;
    if (c < res) {
      res = c;
      ans = i;
    }
  }
  cout << res << endl;
  for (j = 1, k = ans; j <= n; ++j, k += K)
    if (a[j] > k)
      cout << "-"
           << " " << j << " " << a[j] - k << endl;
    else if (a[j] < k)
      cout << "+"
           << " " << j << " " << k - a[j] << endl;
}
