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>
void print(int *aa, int n, int a, int k) {
  int i;
  for (i = 0; i < n; i++) {
    if (aa[i] != a) {
      printf("%c ", aa[i] < a ? '+' : '-');
      printf("%d ", i + 1);
      printf("%d ", aa[i] > a ? aa[i] - a : a - aa[i]);
      printf("\n");
    }
    a += k;
  }
}
int solve(int *aa, int n, int a, int k) {
  int i, cnt;
  cnt = 0;
  for (i = 0; i < n; i++) {
    if (aa[i] != a) cnt++;
    a += k;
  }
  return cnt;
}
int main() {
  static int aa[1000];
  int n, k, i, a, min, amin;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &aa[i]);
  min = n + 1;
  amin = -1;
  for (a = 1; a <= 1000; a++) {
    int cnt = solve(aa, n, a, k);
    if (min > cnt) {
      amin = a;
      min = cnt;
    }
  }
  printf("%d\n", min);
  print(aa, n, amin, k);
  return 0;
}
