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 long double EPS = 1e-9;
const int INF = (int)1e9;
const int N = (int)1e5 + 5;
const long long M = (int)1e9 + 7;
int a[N], b[N];
vector<pair<int, int> > ans, now;
int main() {
  int i, k, v, cur, n, j, p1, p2, f1, f2, ans_l = INF;
  cin >> n >> k;
  for (i = 0; i < n; i++) cin >> a[i];
  for (cur = 0; cur < 1001; cur++) {
    if (!cur) continue;
    now.clear();
    if (a[0] != cur) now.push_back(make_pair(cur - a[0], 0));
    for (j = 1; j < n; j++)
      if (a[j] != cur + j * k) now.push_back(make_pair(cur + j * k - a[j], j));
    if ((int)now.size() < ans_l) ans = now, ans_l = (int)now.size();
  }
  int sz = ans.size();
  cout << sz << endl;
  for (i = 0; i < sz; i++) {
    cur = ans[i].first;
    if (cur < 0)
      cout << "- ";
    else
      cout << "+ ";
    cout << ans[i].second + 1 << ' ' << abs(cur) << endl;
  }
  return 0;
}
