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 argc, char const *argv[]) {
  long long int N, K, zz;
  cin >> N >> K;
  long long int res = N, temp, p = 1, no[N], z;
  for (int i = 0; i < N; i++) cin >> no[i];
  z = 0;
  for (int i = 0; i < N; ++i) {
    p = 0 - i;
    zz = 0;
    temp = 0;
    for (int j = 0; j < N; j++) {
      if (i == j) {
        p++;
        continue;
      }
      if (no[i] + p * K <= 0) {
        zz = 1;
        break;
      }
      if (no[j] != no[i] + p * K) temp++;
      p++;
    }
    if (zz == 0 && temp < res) {
      res = temp;
      z = i;
    }
  }
  cout << res << "\n";
  p = 0 - z;
  for (int i = 0; i < N; ++i) {
    if (res == 0) break;
    if (i == z) {
      p++;
      continue;
    }
    if (no[i] > no[z] + p * K) {
      cout << "- " << i + 1 << " " << no[i] - no[z] - p * K << "\n";
      res--;
    } else if (no[i] < no[z] + p * K) {
      cout << "+ " << i + 1 << " " << no[z] + p * K - no[i] << "\n";
      res--;
    }
    p++;
  }
  return 0;
}
