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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int INF = (1LL << 29);
using namespace std;
template <typename T>
string toString(T t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n, 0);
  for (int i = 0; i < n; i++) cin >> a[i];
  int s = n, sol = 0;
  for (int i = 0; i < n; i++) {
    int curr = 0;
    for (int j = i - 1; j >= 0; j--) {
      if (a[j] - a[i] != (j - i) * k) curr++;
    }
    for (int j = i + 1; j < n; j++) {
      if (a[i] - a[j] != (i - j) * k) curr++;
    }
    bool valid = true;
    for (int j = 0; j < n; j++) {
      if (a[i] + (j - i) * k <= 0) valid = false;
    }
    if (valid && curr < s) {
      s = curr;
      sol = i;
    }
  }
  cout << s << endl;
  for (int j = 0; j < n; j++) {
    int diff = a[j] - a[sol] - (j - sol) * k;
    if (diff < 0) {
      cout << "+ ";
      diff = -diff;
    } else if (diff > 0) {
      cout << "- ";
    }
    if (diff != 0) cout << j + 1 << " " << diff << endl;
  }
  return 0;
}
