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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int a[1010], ccc;
map<int, int> mp;
int main() {
  int n, k;
  while (cin >> n >> k) {
    mp.clear();
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) a[i] -= (i - 1) * k;
    for (int i = 1; i <= n; i++) mp[a[i]]++;
    int ss = 0x3f3f3f3f, tt = 0;
    for (int i = 1; i <= n; i++)
      if (mp[a[i]] > tt && a[i] > 0) tt = mp[a[i]], ss = a[i];
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (a[i] != ss) cnt++;
    cout << cnt << endl;
    for (int i = 1; i <= n; i++)
      if (a[i] != ss) {
        if (a[i] > ss)
          cout << "- ";
        else
          cout << "+ ";
        cout << i << " " << abs(a[i] - ss) << endl;
      }
  }
  return 0;
}
