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 MAX(long long int a, long long int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
int MIN(long long int a, long long int b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
int gcd(long long int a, long long int b) {
  long long int c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return b;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int a[1005];
  int b[1005];
  int i;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int no = 0;
  int save = 1;
  int min = 10000;
  i = 1;
  while (1) {
    if (i == n + 1) {
      break;
    }
    int j;
    for (j = 1; j <= n; j++) {
      b[j] = a[j];
    }
    no = 0;
    for (j = i - 1; j >= 1; j--) {
      if (b[j + 1] - b[j] != k) {
        b[j] = b[j + 1] - k;
      }
      if (b[j] != a[j]) {
        no++;
      }
    }
    for (j = i + 1; j <= n; j++) {
      if (b[j] - b[j - 1] != k) {
        b[j] = b[j - 1] + k;
      }
      if (b[j] != a[j]) {
        no++;
      }
    }
    int ch = 0;
    for (j = 1; j <= n; j++) {
      if (b[j] < 1) {
        ch = 1;
        break;
      }
    }
    if (ch == 1) {
      i++;
      continue;
    }
    if (no < min) {
      min = no;
      save = i;
    }
    i++;
  }
  if (min == 0) {
    cout << "0" << endl;
  } else {
    for (i = 1; i <= n; i++) {
      b[i] = a[i];
    }
    int j;
    i = save;
    for (j = i - 1; j >= 1; j--) {
      if (b[j + 1] - b[j] != k) {
        b[j] = b[j + 1] - k;
      }
    }
    for (j = i + 1; j <= n; j++) {
      if (b[j] - b[j - 1] != k) {
        b[j] = b[j - 1] + k;
      }
    }
    cout << min << endl;
    for (i = 1; i <= n; i++) {
      if (a[i] > b[i]) {
        printf("- %d %d\n", i, a[i] - b[i]);
      } else if (a[i] < b[i]) {
        printf("+ %d %d\n", i, b[i] - a[i]);
      }
    }
  }
  return 0;
}
