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;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
const int mod = 1e9 + 7;
const int mxn = 1e5 + 9;
const long double eps = 1e-9;
long long qpow(long long n, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k >>= 1;
  }
  return ans;
}
int n[1010];
int main() {
  int a, b, c, d, A, w = 0, W;
  scanf("%d %d", &a, &b);
  for (c = 0; c < a; c++) {
    scanf("%d", &n[c]);
  }
  for (c = 0; c < a; c++) {
    if (n[c] - c * b > 0) {
      for (d = 0, A = 0; d < a; d++) {
        if (n[d] == n[c] + b * (d - c)) {
          A++;
        }
      }
      if (A > w) {
        w = A;
        W = c;
      }
    }
  }
  printf("%d\n", a - w);
  for (c = 0; c < a; c++) {
    if (n[c] > n[W] + b * (c - W)) {
      printf("- %d %d\n", c + 1, n[c] - (n[W] + b * (c - W)));
    }
    if (n[c] < n[W] + b * (c - W)) {
      printf("+ %d %d\n", c + 1, n[W] + b * (c - W) - n[c]);
    }
  }
}
