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 ABS(int x) {
  if (x < 0) return -x;
  return x;
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return b / gcd(a, b) * a; }
int n, k, a[1001];
int calc(int idx) {
  int s = 0;
  for (int i = 0; i < n; ++i) {
    if (i == idx) continue;
    int must = a[idx] - (idx - i) * k;
    if (must < 1) return 1 << 30;
    if (must == a[i]) continue;
    ++s;
  }
  return s;
}
void print(int idx) {
  for (int i = 0; i < n; ++i) {
    if (i == idx) continue;
    int must = a[idx] - (idx - i) * k;
    if (must == a[i]) continue;
    if (must > a[i])
      cout << "+ " << i + 1 << " " << must - a[i] << endl;
    else
      cout << "- " << i + 1 << " " << a[i] - must << endl;
  }
}
int main() {
  cin >> n >> k;
  for (int(i) = 0; i < int(n); ++i) cin >> a[i];
  int res = 1 << 20;
  int idx;
  for (int(i) = 0; i < int(n); ++i) {
    int get = calc(i);
    if (get < res) {
      res = get;
      idx = i;
    }
  }
  cout << res << endl;
  print(idx);
  cin >> res;
}
