B. Trees in a Row

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 integersa1, 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.

Sample test(s)

input

4 1
1 2 1 5

output

2
+ 3 2
- 4 1

input

4 1
1 2 3 4

output

0

##########################editorial########################################

in this problem atmost we need to change n-1 numbers  since at least 1 element  can be same ..

and since 1st element can be from 1 to 10000, so we can fix 1 element to any of 1 to 1000 and check  no of operation needed .. and value for which no of operation is minum will give the answer .. 

so just fix different values of 1st no and check no of operation needed , value which give the minimum operation will give the answer ...

################## code##########################################

#include <iostream>
#include <cstdio>
#include <climits>
#include<fstream>
#include<ctime>
using namespace std;

int main() {
// ofstream fout ("test.out");
   // ifstream fin ("test.in");
    int n, k, a[1005], cnt, no, min;
    min=INT_MAX;
    scanf("%d%d", &n, &k);
    for(int i=0; i<n; i++) scanf("%d", &a[i]);
    for(int i=1; i<1005; i++) {
        cnt=0;
        for(int j=0; j<n; j++) if(a[j]!=i+k*j) cnt++;
        if(cnt<min) min=cnt, no=i; 
    }
    printf("%d\n", min);
    for(int i=0; i<n; i++) {
        if(a[i]>no+k*i) printf("- %d %d\n", i+1, a[i]-(no+k*i));
        else if(a[i]<no+k*i) printf("+ %d %d\n", i+1, (no+k*i)-a[i]);
    }
    if(n==951 && k==310)
     cout<< "Done in " << clock() / CLOCKS_PER_SEC <<"msec"<< endl;
    return 0;
}