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 a[1010];
int visited[1010];
vector<vector<int> > vv;
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  vv.clear();
  int i, j;
  for (i = 1; i <= n; ++i) cin >> a[i];
  for (i = 1; i <= n; ++i) {
    do {
      if (false)
        printf(
            "start i loop, i:%d, a[i]:%d, k:%d, visited[i]:%d"
            "\n",
            i, a[i], k, visited[i]);
    } while (false);
    if (a[i] - (i - 1) * k <= 0 || visited[i] == 1) continue;
    vector<int> tv;
    tv.clear();
    tv.push_back(i);
    visited[i] = 1;
    do {
      if (false)
        printf(
            "begin i:%d"
            "\n",
            i);
    } while (false);
    for (j = i + 1; j <= n; ++j) {
      if (visited[j] == 0 && a[j] == a[i] + (j - i) * k) {
        tv.push_back(j);
        visited[j] = 1;
        do {
          if (false)
            printf(
                "catch j:%d"
                "\n",
                j);
        } while (false);
      }
    }
    vv.push_back(tv);
  }
  do {
    if (false)
      printf(
          "vv.size:%d"
          "\n",
          vv.size());
  } while (false);
  int ri, ms;
  ri = 0;
  ms = -1;
  for (i = 0; i < (int)vv.size(); ++i) {
    do {
      if (false)
        printf(
            "vv[%d].size:%d, ms:%d"
            "\n",
            i, vv[i].size(), ms);
    } while (false);
    if ((int)vv[i].size() > ms) {
      ms = (int)vv[i].size();
      do {
        if (false)
          printf(
              "got ms:%d"
              "\n",
              ms);
      } while (false);
      ri = i;
    } else {
      do {
        if (false)
          printf(
              "oooops"
              "\n");
      } while (false);
    }
  }
  do {
    if (false)
      printf(
          "ri:%d"
          "\n",
          ri);
  } while (false);
  cout << n - (int)vv[ri].size() << endl;
  int h = vv[ri][0];
  int delta;
  do {
    if (false)
      printf(
          "h:%d"
          "\n",
          h);
  } while (false);
  for (i = 1; i <= n; ++i) {
    delta = a[i] - (a[h] + (i - h) * k);
    do {
      if (false)
        printf(
            "delta:%d"
            "\n",
            delta);
    } while (false);
    if (delta > 0) {
      cout << "- " << i << " " << delta << endl;
    } else if (delta < 0) {
      cout << "+ " << i << " " << -delta << endl;
    } else {
    }
  }
  return 0;
}
