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 N = 1005;
int n, k, a[N], b[N], fin[N], res, ans;
const int inf = 0x3f3f3f3f;
queue<int> A, B;
int main() {
  scanf("%d%d", &n, &k);
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]), mx = max(a[i], mx);
  }
  ans = inf;
  for (int beg = 1; beg <= mx; ++beg) {
    res = 0;
    for (int i = 1, j = beg; i <= n; ++i, j += k) {
      b[i] = j - a[i];
      if (b[i]) ++res;
    }
    if (res < ans) {
      ans = res;
      for (int j = 1; j <= n; ++j) fin[j] = b[j];
    }
  }
  printf("%d\n", ans);
  for (int i = 1; i <= n; ++i)
    if (fin[i] > 0)
      printf("+ %d %d\n", i, fin[i]);
    else if (fin[i] < 0)
      printf("- %d %d\n", i, -fin[i]);
  return 0;
}
