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 n, k, a[10005];
int solve(int i, int height, vector<pair<bool, pair<int, int> > > &out) {
  if (i == n) return 0;
  if (height + k == a[i]) return solve(i + 1, height + k, out);
  if (height + k > a[i])
    out.push_back(make_pair(1, make_pair(i, height + k - a[i])));
  else
    out.push_back(make_pair(0, make_pair(i, a[i] - (height + k))));
  return 1 + solve(i + 1, height + k, out);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  vector<pair<bool, pair<int, int> > > out, lout;
  int sol = solve(1, a[0], out);
  for (int i = 1; i <= 1000; ++i)
    if (i != a[0]) {
      lout.clear();
      if (i > a[0])
        lout.push_back(make_pair(1, make_pair(0, i - a[0])));
      else
        lout.push_back(make_pair(0, make_pair(0, a[0] - i)));
      int lsol = 1 + solve(1, i, lout);
      if (sol > lsol) sol = lsol, out = lout;
    }
  printf("%d\n", sol);
  for (int i = 0; i < sol; ++i) {
    printf("%c ", (out[i].first) ? '+' : '-');
    printf("%d %d\n", out[i].second.first + 1, out[i].second.second);
  }
  return 0;
}
