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;
template <class T>
string toStr(const T &x) {
  stringstream s;
  s << x;
  return s.str();
}
template <class T>
int toInt(const T &x) {
  stringstream s;
  s << x;
  int r;
  s >> r;
  return r;
}
int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << 0;
    return 0;
  }
  int v[n];
  for (int i = 0; i < n; ++i) cin >> v[i];
  int best = 0;
  int mmax = -1;
  for (int i = 0; i < n; ++i) {
    int tmp = 0;
    int f = true;
    for (int j = i + 1, kk = v[j - 1] + k; j < n; ++j, kk += k) {
      if (kk == v[j]) tmp++;
    }
    for (int j = i - 1, kk = v[j + 1] - k; j >= 0; --j, kk -= k) {
      if (kk == v[j]) tmp++;
      if (kk <= 0) f = false;
    }
    if (tmp > mmax && f) {
      best = i;
      mmax = tmp;
    }
  }
  vector<string> ans;
  for (int j = best + 1, kk = v[j - 1] + k; j < n; ++j, kk += k) {
    if (kk != v[j]) {
      if (kk > v[j]) {
        ans.push_back("+ " + toStr(j + 1) + " " + toStr(kk - v[j]));
      } else {
        ans.push_back("- " + toStr(j + 1) + " " + toStr(v[j] - kk));
      }
    }
  }
  for (int j = best - 1, kk = v[j + 1] - k; j >= 0; --j, kk -= k) {
    if (kk != v[j]) {
      if (kk > v[j]) {
        ans.push_back("+ " + toStr(j + 1) + " " + toStr(kk - v[j]));
      } else {
        ans.push_back("- " + toStr(j + 1) + " " + toStr(v[j] - kk));
      }
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}
