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 int maxn = 1005;
int tree[maxn];
int n, k;
int i_b = 0;
int min_opr = 99999999;
int checktree(int i) {
  int t_opr = 0;
  int t_b = tree[i] - k * i;
  for (int i = 1; i <= n; i++)
    if (k * i + t_b != tree[i]) t_opr++;
  if (min_opr > t_opr && (k + t_b >= 1)) {
    min_opr = t_opr;
    i_b = t_b;
  }
  return t_opr;
}
int main() {
  memset(tree, 0, sizeof(tree));
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tree[i]);
  }
  for (int i = 1; i <= n; i++) {
    checktree(i);
  }
  cout << min_opr << endl;
  for (int i = 1; i <= n; i++) {
    int dis = k * i + i_b - tree[i];
    if (dis < 0)
      cout << "- " << i << " " << abs(dis) << endl;
    else if (dis > 0)
      cout << "+ " << i << " " << abs(dis) << endl;
  }
}
