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 int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y / 2;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  long long int a[n + 1];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<pair<long long int, long long int> > vp;
  for (long long int i = 1; i <= 1000; i++) {
    vector<long long int> v;
    v.push_back(i);
    for (long long int j = 1; j < n; j++) {
      v.push_back(v[j - 1] + k);
    }
    long long int c = 0;
    for (long long int j = 0; j < n; j++)
      if (v[j] != a[j]) c++;
    vp.push_back({c, i});
  }
  sort(vp.begin(), vp.end());
  cout << vp[0].first << "\n";
  vector<long long int> ans;
  ans.push_back(vp[0].second);
  for (long long int i = 1; i < n; i++) ans.push_back(ans[i - 1] + k);
  for (long long int i = 0; i < n; i++) {
    if (a[i] != ans[i]) {
      if (ans[i] - a[i] > 0)
        cout << "+"
             << " " << i + 1 << " " << abs(ans[i] - a[i]) << "\n";
      else
        cout << "-"
             << " " << i + 1 << " " << abs(ans[i] - a[i]) << "\n";
    }
  }
}
int main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
