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>
int fix[1001];
int change[1001];
int main() {
  int i, n, j, k, max = 0, min = 1001, in[1001], a, pos, expected, mins = 0;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%d", &in[i]);
    if (in[i] > max) {
      max = in[i];
    }
  }
  for (i = 1; i <= max; i++) {
    for (j = 1; j <= n; j++) {
      expected = i + ((j - 1) * k);
      if (in[j] != expected) {
        fix[i]++;
      }
    }
    if (fix[i] == 0) {
      a = i;
      min = 0;
      break;
    } else if (fix[i] < min) {
      min = fix[i];
      a = i;
    }
  }
  for (i = 1; i <= n; i++) {
    expected = a + ((i - 1) * k);
    if (in[i] != expected) {
      change[i] = expected - in[i];
      mins++;
    }
  }
  printf("%d\n", mins);
  for (i = 1; i < 1001; i++) {
    if (change[i] > 0) {
      printf("+ %d %d\n", i, change[i]);
    } else if (change[i] < 0) {
      printf("- %d %d\n", i, (change[i] * -1));
    }
  }
}
