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:64000000")
using namespace std;
int a[10000], b[10000] = {0};
bool used[10];
vector<int> path;
long long dist[1000];
int kol = 0;
int res = 100000;
string s, s1, s2;
long long t, i, j = 1, poss, posg, n = 0, start, iter, m, z, k = 1, r, pr = 0,
                maxk = 0, inf = 30000, xf, yf, xs, ys;
int kol1 = 0, kol2 = 0;
long double fact = 1;
int main() {
  long long now, x1 = 0, res1 = 0, sum1 = 0, k, v, sum2 = 0, proizv1 = 1, y1,
                 x2 = 0, maxz = -1, y2, x;
  double rast = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (j = 1; j <= 1000; j++) {
    now = 0;
    for (i = 1; i <= n; i++) {
      if (a[i] != j + (i - 1) * k) now++;
    }
    if (now < res) {
      res = now;
      start = j;
    }
  }
  cout << res << endl;
  for (i = 1; i <= n; i++) {
    if (a[i] > start + (i - 1) * k)
      cout << '-' << ' ' << i << ' ' << a[i] - start - (i - 1) * k << endl;
    else {
      if (a[i] < start + (i - 1) * k)
        cout << '+' << ' ' << i << ' ' << start + (i - 1) * k - a[i] << endl;
    }
  }
  return 0;
}
