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 di[4] = {1, 0, -1, 0};
int dj[4] = {0, 1, 0, -1}, inf = 2147483647;
const long long ll = 9223372036854775807;
struct node {
  char a;
  int b, c;
};
int main() {
  int a, b;
  while (cin >> a >> b) {
    int A[1024] = {};
    for (int i = 0; i < a; i++) {
      cin >> A[i];
    }
    vector<node> ans;
    int maxr = inf, id = 0;
    for (int i = 1; i <= 1000; i++) {
      int sum = 0;
      int now = i;
      for (int j = 0; j < a; j++) {
        if (A[j] != now) {
          sum++;
        }
        now += b;
      }
      if (sum < maxr) {
        maxr = sum;
        id = i;
      }
    }
    int now = id;
    for (int i = 0; i < a; i++) {
      if (A[i] > now) {
        node tmp;
        tmp.a = '-', tmp.b = i + 1, tmp.c = A[i] - now;
        ans.push_back(tmp);
      }
      if (A[i] < now) {
        node tmp;
        tmp.a = '+', tmp.b = i + 1, tmp.c = now - A[i];
        ans.push_back(tmp);
      }
      now += b;
    }
    cout << ans.size() << '\n';
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i].a << " " << ans[i].b << " " << ans[i].c << '\n';
    }
  }
  return 0;
}
