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 Fun(int *data, int i, int j, int K) {
  if (j < i) {
    return data[i] - (i - j) * K;
  } else {
    return data[i] + (j - i) * K;
  }
}
int main() {
  int N, K, i, j, Min, Minid;
  int A[1000];
  scanf("%d %d", &N, &K);
  for (i = 0; i < N; ++i) {
    scanf("%d", &A[i]);
  }
  Min = 0x7fffffff;
  for (i = 0; i < N; ++i) {
    int Cnt = 0;
    for (j = 0; j < N; ++j) {
      if (i == j) {
        continue;
      }
      if (A[j] != Fun(A, i, j, K)) {
        ++Cnt;
        if (Fun(A, i, j, K) <= 0) {
          Cnt += 2 * N;
          break;
        }
      }
    }
    if (Cnt < Min) {
      Min = Cnt;
      Minid = i;
    }
  }
  printf("%d\n", Min);
  for (i = 0; i < N; ++i) {
    int dst = Fun(A, Minid, i, K);
    if (A[i] > dst) {
      printf("- %d %d\n", i + 1, A[i] - dst);
    } else if (A[i] < dst) {
      printf("+ %d %d\n", i + 1, dst - A[i]);
    }
  }
  return 0;
}
