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 a[1002], visited[1002];
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  int i;
  for (i = 0; i < N; i++) {
    scanf("%d", &a[i]);
  }
  int maxi = 0, indx, j;
  for (i = 0; i < N; i++) {
    int count = 0;
    if (visited[i] == 0) {
      if (a[i] - i * K <= 0) continue;
      for (j = i; j < N; j++) {
        int p = (j - i);
        if (a[j] == p * K + a[i]) {
          count++;
          visited[j] = 1;
        }
      }
    }
    if (count > maxi) {
      maxi = count;
      indx = i;
    }
  }
  printf("%d\n", N - maxi);
  for (i = 0; i < N; i++) {
    int exp = a[indx] + (i - indx) * K;
    if (a[i] == exp) continue;
    if (a[i] > exp) {
      printf("- %d %d\n", i + 1, a[i] - exp);
    } else {
      printf("+ %d %d\n", i + 1, exp - a[i]);
    }
  }
}
