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 main() {
  int a[1001], b[1001], c[1001];
  int n, m;
  while (cin >> n >> m) {
    memset(c, 0, sizeof(c));
    memset(b, 0, sizeof(b));
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    for (int i = 0; i < n; i++) {
      int count = 0;
      for (int j = 0; j < i; j++) {
        b[j] = a[i] - m * (i - j);
      }
      for (int j = i; j < n; j++) {
        b[j] = a[i] + m * (j - i);
      }
      for (int j = 0; j < n; j++) {
        if (b[j] != a[j]) {
          count++;
        }
        if (b[0] <= 0) {
          count = 99999999;
        }
      }
      c[i] = count;
    }
    int minn = 99999999;
    int k;
    for (int i = 0; i < n; i++) {
      if (c[i] < minn) {
        minn = c[i];
        k = i;
      }
    }
    for (int j = 0; j < k; j++) {
      b[j] = a[k] - m * (k - j);
    }
    for (int j = k; j < n; j++) {
      b[j] = a[k] + m * (j - k);
    }
    cout << minn << endl;
    for (int i = 0; i < n; i++) {
      int num;
      if (b[i] != a[i]) {
        num = b[i] - a[i];
        if (num < 0) {
          cout << "- " << i + 1 << " " << -num << endl;
        } else {
          cout << "+ " << i + 1 << " " << num << endl;
        }
      }
    }
  }
  return 0;
}
