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[10000], b[10000], c[1001][10000];
int tot, Min = 99999999, posi;
int main() {
  int i, k, n, xx;
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (xx = 1; xx <= 1000; xx++) {
    b[1] = xx;
    for (i = 2; i <= n; i++) {
      b[i] = b[i - 1] + k;
      c[xx][i] = b[i] - a[i];
    }
    c[xx][1] = b[1] - a[1];
    for (i = 1; i <= n; i++) {
      if (c[xx][i] != 0) {
        tot++;
      }
    }
    if (tot < Min) {
      Min = tot;
      posi = xx;
    }
    tot = 0;
  }
  if (Min == 0) {
    printf("0");
    return 0;
  }
  cout << Min << endl;
  for (i = 1; i <= n; i++) {
    if (c[posi][i] == 0)
      continue;
    else if (c[posi][i] < 0) {
      printf("- %d %d\n", i, -c[posi][i]);
    } else if (c[posi][i] > 0) {
      printf("+ %d %d\n", i, c[posi][i]);
    }
  }
  return 0;
}
