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 i, j, k, l, m, n, o, p, r, s, t;
  vector<int> strom;
  scanf("%d", &n);
  scanf("%d", &k);
  strom.resize(n);
  for (i = 0; i < n; i++) {
    scanf("%d", &p);
    strom[i] = p;
  }
  m = n + 5;
  r = 0;
  for (i = -k + 1; i <= 1000; i++) {
    p = 0;
    for (j = 0; j < n; j++) {
      if (strom[j] != i + k * (j + 1)) {
        p++;
      }
    }
    if (p < m) {
      m = p;
      r = i;
    }
  }
  printf("%d\n", m);
  for (i = 0; i < n; i++) {
    if (strom[i] < r + k * (i + 1)) {
      printf("+ %d %d\n", i + 1, r + k * (i + 1) - strom[i]);
    }
    if (strom[i] > r + k * (i + 1)) {
      printf("- %d %d\n", i + 1, strom[i] - (r + k * (i + 1)));
    }
  }
  return 0;
}
