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>
const int INF = 1e9;
const int N = 10500;
const double PI = 3.1415926535897932384626433832795;
using namespace std;
int a[N], k, n, mx = -1;
int main() {
  ios_base ::sync_with_stdio();
  int c, i;
  scanf("%d %d", &n, &k);
  for (c = 1; c <= n; c++) {
    scanf("%d", &a[c]);
  }
  int pos = -1;
  for (c = 1; c <= 1000; c++) {
    int as = 0, now = c;
    for (i = 1; i <= n; i++) {
      if (a[i] == now) as++;
      now += k;
    }
    if (as > mx) {
      mx = as;
      pos = c;
    }
  }
  cout << n - mx << endl;
  int now = pos;
  for (c = 1; c <= n; c++) {
    if (a[c] != now) {
      if (a[c] < now)
        cout << "+"
             << " " << c << " " << now - a[c] << endl;
      else
        cout << "- " << c << " " << a[c] - now << endl;
    }
    now += k;
  }
  return 0;
}
