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 num[1005], vis[1005];
int main() {
  int n, k, step = 9999, in;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  for (int i = 1; i <= n; i++) {
    int t = i, ans = 0, temp = 1;
    vis[t] = num[t];
    while (t != 1) {
      if (num[t - 1] + k != vis[t]) ans++;
      vis[t - 1] = vis[t] - k;
      if (vis[t - 1] <= 0) {
        temp = 0;
        break;
      }
      t--;
    }
    if (!temp) continue;
    t = i;
    while (t != n) {
      if (vis[t] + k != num[t + 1]) ans++;
      vis[t + 1] = vis[t] + k;
      t++;
    }
    if (ans < step) {
      step = ans;
      in = i;
    }
  }
  printf("%d\n", step);
  int t = in;
  while (t != 1) {
    int ans = num[t - 1] + k - num[t];
    if (ans > 0) printf("- %d %d\n", t - 1, ans);
    if (ans < 0) printf("+ %d %d\n", t - 1, -ans);
    num[t - 1] = num[t] - k;
    t--;
  }
  t = in;
  while (t != n) {
    int ans = num[t] + k - num[t + 1];
    if (ans > 0) printf("+ %d %d\n", t + 1, ans);
    if (ans < 0) printf("- %d %d\n", t + 1, -ans);
    num[t + 1] = num[t] + k;
    t++;
  }
  return 0;
}
