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 n, m, k;
int a[100010];
struct node {
  int x, id;
} p[100010];
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int ans = 99999999;
    int pos = 1;
    for (int i = 1; i <= n; i++) {
      bool flag = false;
      int cnt = 0;
      for (int j = i; j >= 1; j--) {
        if (a[i] - (i - j) * k == a[j]) continue;
        cnt++;
        if (a[i] - (i - j) * k <= 0) {
          flag = true;
          break;
        }
      }
      if (flag) continue;
      int cnt1 = 0;
      for (int j = i; j <= n; j++) {
        if (a[i] + (j - i) * k == a[j]) continue;
        cnt1++;
      }
      if (cnt1 + cnt < ans) {
        ans = cnt1 + cnt;
        pos = i;
      }
    }
    printf("%d\n", ans);
    for (int j = 1; j <= pos; j++) {
      if (a[j] + (pos - j) * k == a[pos]) continue;
      if (a[j] < a[pos] - (pos - j) * k) {
        printf("+ %d %d\n", j, a[pos] - (pos - j) * k - a[j]);
      } else {
        printf("- %d %d\n", j, a[j] - (a[pos] - (pos - j) * k));
      }
    }
    for (int j = pos; j <= n; j++) {
      if (a[pos] + (j - pos) * k == a[j]) continue;
      if (a[j] < a[pos] + (j - pos) * k) {
        printf("+ %d %d\n", j, a[pos] + (j - pos) * k - a[j]);
      } else {
        printf("- %d %d\n", j, a[j] - (a[pos] + (j - pos) * k));
      }
    }
  }
  return 0;
}
