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;
const int MAX = 101;
template <class T, class U>
void convert(T &t, U &u) {
  stringstream ss;
  ss << t;
  ss >> u;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> tb;
  for (int(i) = 0; (i) < (n); (i)++) {
    int tmp;
    cin >> tmp;
    tb.push_back(tmp);
  }
  int ans = 1001;
  int idx = -1;
  for (int(i) = 0; (i) < (tb.size()); (i)++) {
    int c = 0;
    int prev1 = tb[i], prev2 = tb[i];
    for (int j = i + 1; j < tb.size(); j++) {
      int t = (prev1 + k) - tb[j];
      if (t > 0 || t < 0) c++;
      prev1 += k;
    }
    bool f = true;
    for (int j = i - 1; j >= 0; j--) {
      int t = (prev2 - k) - tb[j];
      if (t > 0 || t < 0) c++;
      prev2 -= k;
      if (prev2 <= 0) {
        f = false;
        break;
      }
    }
    if (!f) continue;
    if (ans > c) {
      idx = i;
      ans = c;
    }
  }
  vector<string> ans1;
  int prev1 = tb[idx], prev2 = tb[idx];
  for (int j = idx + 1; j < tb.size(); j++) {
    int t = (prev1 + k) - tb[j];
    if (t > 0) {
      string tt;
      tt += "+ ";
      int l = j + 1;
      ostringstream ss, sss;
      ss << l;
      sss << t;
      tt = string("+ " + ss.str() + " " + sss.str());
      ans1.push_back(tt);
    } else if (t < 0) {
      string tt;
      tt += "- ";
      int l = j + 1;
      ostringstream ss, sss;
      ss << l;
      t = abs(t);
      sss << t;
      tt = string("- " + ss.str() + " " + sss.str());
      ans1.push_back(tt);
    }
    prev1 += k;
  }
  for (int j = idx - 1; j >= 0; j--) {
    int t = (prev2 - k) - tb[j];
    if (t > 0) {
      string tt;
      tt += "+ ";
      ostringstream ss, sss;
      int l = j + 1;
      ss << l;
      sss << t;
      tt = string("+ " + ss.str() + " " + sss.str());
      ans1.push_back(tt);
    } else if (t < 0) {
      string tt;
      tt += "- ";
      int l = j + 1;
      ostringstream ss, sss;
      ss << l;
      t = abs(t);
      sss << t;
      tt = string("- " + ss.str() + " " + sss.str());
      ans1.push_back(tt);
    }
    prev2 -= k;
  }
  cout << ans1.size() << endl;
  for (int(i) = 0; (i) < (ans1.size()); (i)++) {
    cout << ans1[i] << endl;
  }
  return 0;
}
