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, m, k, l, r, t, temp, cnt, ans, res, x, sum;
int a[1001];
int b[1001];
int main() {
  res = 1000000;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= 1000; i++) {
    b[0] = i;
    for (int j = 1; j < n; j++) {
      b[j] = b[j - 1] + k;
    }
    cnt = 0;
    for (int j = 0; j < n; j++) {
      if (a[j] != b[j]) cnt++;
    }
    if (cnt < res) {
      ans = i;
      res = cnt;
    }
  }
  cout << res << endl;
  b[0] = ans;
  for (int j = 1; j < n; j++) {
    b[j] = b[j - 1] + k;
  }
  for (int j = 0; j < n; j++) {
    if (a[j] > b[j]) {
      cout << "- " << j + 1 << " " << a[j] - b[j] << endl;
    }
    if (a[j] < b[j]) {
      cout << "+ " << j + 1 << " " << -(a[j] - b[j]) << endl;
    }
  }
  return 0;
}
