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 aa[2005];
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> aa[i];
  }
  int minx = 2005;
  int best = 0;
  for (int i = 0; i < n; ++i) {
    int tmp = 0;
    int can = 1;
    for (int t = 0; t < i; ++t) {
      if (aa[t] != aa[i] - (i - t) * k) tmp++;
      if ((aa[i] - (i - t) * k) <= 0) {
        can = 0;
        break;
      }
    }
    if (!can) continue;
    for (int t = i + 1; t < n; ++t) {
      if (aa[t] != aa[i] + (t - i) * k) tmp++;
    }
    if (tmp < minx) {
      minx = tmp;
      best = i;
      if (minx == 0) {
        cout << 0 << endl;
        return 0;
      }
    }
  }
  cout << minx << endl;
  for (int i = 0; i < n; ++i) {
    int tar = aa[best] + (i - best) * k;
    if (aa[i] < tar) {
      cout << '+' << ' ';
      cout << i + 1 << ' ' << tar - aa[i] << endl;
    } else if (aa[i] > tar) {
      cout << '-' << ' ' << i + 1 << ' ' << aa[i] - tar << endl;
    }
  }
  return 0;
}
