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 main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int N, k;
  cin >> N >> k;
  int A[N];
  for (int i = 0; i < N; ++i) cin >> A[i];
  int ans = N - 1;
  int M = A[0];
  for (int a = 1; a <= 1000; ++a) {
    int mins = 0;
    if (a != A[0]) ++mins;
    for (int i = 1; i < N; ++i)
      if (A[i] != a + (k * i)) ++mins;
    if (ans > mins) ans = mins, M = a;
  }
  cout << ans;
  for (int i = 0; i < N; ++i) {
    if (A[i] == M + k * i) continue;
    cout << '\n';
    if (A[i] > M + k * i)
      cout << '-';
    else
      cout << '+';
    cout << ' ' << i + 1 << ' ' << abs(A[i] - (M + k * i));
  }
  return 0;
}
