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;
time_t start = clock();
typedef long long lld;
typedef double lf;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
struct ooo {
  char c;
  int a, b;
} O[1009], O1[1009];
int n, k, z, p, mn = (int)1e9 + 9;
int v[1009];
bool one;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= n; i++) {
    z = 0, one = 0;
    p = v[i] - k;
    for (int h = i - 1; h > 0; h--) {
      if (p < 1) {
        one = 1;
        break;
      }
      if (p != v[h]) {
        if (p > v[h])
          O[z].c = '+', O[z].a = h, O[z++].b = p - v[h];
        else
          O[z].c = '-', O[z].a = h, O[z++].b = v[h] - p;
      }
      p -= k;
    }
    p = v[i] + k;
    if (!one) {
      for (int h = i + 1; h <= n; h++) {
        if (p != v[h]) {
          if (p > v[h])
            O[z].c = '+', O[z].a = h, O[z++].b = p - v[h];
          else
            O[z].c = '-', O[z].a = h, O[z++].b = v[h] - p;
        }
        p += k;
      }
    }
    if (mn > z && !one) {
      mn = z;
      for (int h = 0; h < z; h++) O1[h] = O[h];
    }
  }
  cout << mn << "\n";
  for (int i = 0; i < mn; i++)
    cout << O1[i].c << " " << O1[i].a << " " << O1[i].b << "\n";
  return 0;
}
