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 read() {
  int ret = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && (ch ^ '-')) ch = getchar();
  if (!(ch ^ '-')) f = -1, ch = getchar();
  while (isdigit(ch)) {
    ret = (ret << 3) + (ret << 1) + (ch ^ '0'), ch = getchar();
  }
  return ret * f;
}
const int maxn = 1500;
int n, k;
int a[maxn];
int cnt, tot = maxn;
int ans;
int tt;
struct node {
  int opt, pos, ans;
} p[maxn];
int main() {
  n = read(), k = read();
  for (register int i = 1; i <= n; ++i) {
    a[i] = read();
  }
  for (register int i = 1; i <= n; ++i) {
    cnt = 0;
    for (register int j = i - 1; j >= 1; --j) {
      if (a[i] - (i - j) * k <= 0) {
        cnt = maxn;
        break;
      }
      if (a[j] == a[i] - (i - j) * k) {
        continue;
      } else {
        cnt++;
        if (cnt > tot) break;
      }
    }
    for (register int j = i + 1; j <= n; ++j) {
      if (a[j] == a[i] + (j - i) * k) {
        continue;
      } else {
        cnt++;
        if (cnt > tot) break;
      }
    }
    if (cnt < tot) {
      tot = cnt;
      ans = i;
    }
  }
  if (tot == maxn) {
    for (register int i = 1; i <= n; ++i) {
      if (a[i] > i * k) printf("- %d %d\n", i, abs(a[i] - i * k));
      if (a[i] < i * k) printf("+ %d %d\n", i, abs(a[i] - i * k));
    }
    return 0;
  }
  for (register int j = ans - 1; j >= 1; --j) {
    if (a[j] < a[ans] - (ans - j) * k) {
      p[++tt].opt = 1;
      p[tt].pos = j;
      p[tt].ans = abs(a[ans] - (ans - j) * k - a[j]);
    } else if (a[j] > a[ans] - (ans - j) * k) {
      p[++tt].opt = 0;
      p[tt].pos = j;
      p[tt].ans = abs(a[ans] - (ans - j) * k - a[j]);
    }
  }
  for (register int j = ans + 1; j <= n; ++j) {
    if (a[j] < a[ans] + (j - ans) * k) {
      p[++tt].opt = 1;
      p[tt].pos = j;
      p[tt].ans = abs(a[ans] + (j - ans) * k - a[j]);
    } else if (a[j] > a[ans] + (j - ans) * k) {
      p[++tt].opt = 0;
      p[tt].pos = j;
      p[tt].ans = abs(a[ans] + (j - ans) * k - a[j]);
    }
  }
  printf("%d\n", tt);
  for (register int i = 1; i <= tt; ++i) {
    if (p[i].opt == 0) printf("- %d %d\n", p[i].pos, p[i].ans);
    if (p[i].opt == 1) printf("+ %d %d\n", p[i].pos, p[i].ans);
  }
  return 0;
}
