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 n, k;
  cin >> n >> k;
  bool f = true;
  int a[n + 1];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i) {
      if (a[i] - a[i - 1] != k) {
        f = false;
      }
    }
  }
  if (f) {
    cout << "0" << endl;
  } else {
    int m[n + 1];
    for (int i = 0; i < n; i++) {
      m[i] = 100000;
    }
    for (int i = 0; i < n; i++) {
      int se = i;
      int ctr = 0;
      int nx = a[se] + k;
      for (int j = se + 1; j < n; j++) {
        if (nx != a[j]) {
          ctr++;
        }
        nx += k;
      }
      nx = a[se] - k;
      int pre = 100;
      for (int j = se - 1; j >= 0; j--) {
        pre = nx;
        if (nx != a[j]) {
          ctr++;
        }
        nx -= k;
      }
      if (pre <= 0) {
        ctr = 100000;
      }
      m[i] = ctr;
    }
    int indx = -1;
    int m2 = 100000;
    for (int i = 0; i < n; i++) {
      if (m[i] < m2) {
        m2 = m[i];
        indx = i;
      }
    }
    cout << m2 << endl;
    int to[n + 1];
    int nx = a[indx] + k;
    for (int j = indx + 1; j < n; j++) {
      to[j] = nx;
      nx += k;
    }
    nx = a[indx] - k;
    for (int j = indx - 1; j >= 0; j--) {
      to[j] = nx;
      nx -= k;
    }
    for (int i = 0; i < n; i++) {
      if (i == indx) {
        continue;
      }
      if (to[i] < a[i]) {
        cout << "- " << i + 1 << " " << a[i] - to[i] << endl;
      } else if (to[i] > a[i]) {
        cout << "+ " << i + 1 << " " << to[i] - a[i] << endl;
      }
    }
  }
  return 0;
}
