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 n, k, a[1002], tmp[1002], ba, cnt, MIN = 10000, ans = -1, K[1002];
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    K[i] = (i - 1) * k;
  }
  for (int i = 1; i <= n; ++i) {
    ba = a[i] - K[i];
    if (ba <= 0) continue;
    cnt = 0;
    for (int j = 1; j <= n; ++j) {
      if (a[j] != ba + K[j]) {
        ++cnt;
      }
      if (cnt >= MIN) {
        cnt = -1;
        break;
      }
    }
    if (cnt != -1) {
      if (cnt < MIN) {
        MIN = cnt;
        ans = i;
      }
    }
  }
  if (MIN == 0) {
    puts("0");
    return 0;
  }
  ba = a[ans] - K[ans];
  for (int i = 1; i <= n; ++i) tmp[i] = ba + K[i];
  printf("%d\n", MIN);
  for (int i = 1; i <= n; ++i)
    if (tmp[i] > a[i])
      printf("+ %d %d\n", i, tmp[i] - a[i]);
    else if (a[i] > tmp[i])
      printf("- %d %d\n", i, a[i] - tmp[i]);
}
