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 on_bit(int x, int pos) {
  x |= (1 << pos);
  return x;
}
int off_bit(int x, int pos) {
  x &= ~(1 << pos);
  return x;
}
bool is_on_bit(int x, int pos) { return ((x & (1 << pos)) != 0); }
int flip_bit(int x, int pos) {
  x ^= (1 << pos);
  return x;
}
int lsb(int x) { return x & (-x); }
int on_bit_all(int x, int pos) {
  x = (1 << pos) - 1;
  return x;
}
const double EPS = 1e-9;
const double PI = 2 * acos(0.0);
const int INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int MOD = 1e9 + 7;
int add(long long a, long long b) { return ((a % MOD) + (b % MOD)) % MOD; }
int sub(long long a, long long b) { return ((a % MOD) - (b % MOD)) % MOD; }
int mult(long long a, long long b) { return ((a % MOD) * (b % MOD)) % MOD; }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int N = 1234;
int a[N];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < (int)n; i++) scanf("%d", a + i);
  int curr_moves = -INF, curr_min_moves = INF;
  vector<pair<char, pair<int, int> > > res;
  for (int i = 1; i < (int)1001; i++) {
    vector<pair<char, pair<int, int> > > u;
    int curr_moves = 0;
    for (int j = 0; j < (int)n; j++) {
      if (a[j] != i + j * k) {
        curr_moves++;
        u.push_back(make_pair(a[j] < i + j * k ? '+' : '-',
                              make_pair(j + 1, abs(a[j] - (i + j * k)))));
      }
    }
    if (curr_moves < curr_min_moves) {
      curr_min_moves = curr_moves;
      res = u;
    }
  }
  printf("%d\n", ((int)(res).size()));
  for (int i = 0; i < (int)((int)(res).size()); i++) {
    printf("%c %d %d\n", res[i].first, res[i].second.first,
           res[i].second.second);
  }
  return 0;
}
