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 a[1001];
int ans[1001];
int n, k;
int main() {
  while (cin >> n >> k) {
    memset(ans, 0, sizeof(ans));
    for (int i = 1; i <= n; i++) cin >> a[i];
    int max = 10000;
    for (int i = 1; i <= n; i++) {
      int s = 0;
      for (int t = 1; t <= n; t++) {
        if (a[i] - a[t] != k * (i - t)) s++;
      }
      if (s < max) {
        if (a[i] - k * (i - 1) <= 0) continue;
        for (int t = 1; t <= n; t++) {
          ans[t] = k * (t - i) - a[t] + a[i];
        }
        max = s;
      }
    }
    cout << max << endl;
    for (int i = 1; i <= n; i++) {
      if (ans[i] > 0) cout << "+ " << i << " " << ans[i] << endl;
      if (ans[i] < 0) cout << "- " << i << " " << -ans[i] << endl;
    }
  }
  return 0;
}
