codeforces402B

Trees in a Row

 CodeForces - 402B 

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 a_i 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), a_i + 1 - a_i = 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, ..., a_n (1 ≤ a_i ≤ 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

sol:数据范围小的可怜，爆枚一个正确节点，n2模拟即可

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=1005;
int n,m,a[N],b[N],Ans[N];
int main()
{
    int i,j,Pos=-1;
    R(n); R(m);
    for(i=1;i<=n;i++) R(a[i]);
    for(i=1;i<=n;i++)
    {
        Ans[i]=0;
        b[i]=a[i];
        for(j=i-1;j>=1;j--) b[j]=b[j+1]-m;
        for(j=i+1;j<=n;j++) b[j]=b[j-1]+m;
        for(j=1;j<=n;j++)
        {
            if(b[j]!=a[j]) Ans[i]++;
            if(b[j]<=0) {Ans[i]=0x3f3f3f3f; break;}
        }
        if((Pos==-1)||(Ans[i]<Ans[Pos])) Pos=i;
    }
    Wl(Ans[Pos]);
    b[Pos]=a[Pos];
    for(i=Pos-1;i>=1;i--) b[i]=b[i+1]-m;
    for(i=Pos+1;i<=n;i++) b[i]=b[i-1]+m;
    for(i=1;i<=n;i++) if(a[i]!=b[i])
    {
        if(a[i]<b[i])
        {
            putchar('+'); putchar(' '); W(i); Wl(b[i]-a[i]);
        }
        else
        {
            putchar('-'); putchar(' '); W(i); Wl(a[i]-b[i]);
        }
    }
    return 0;
}
/*
Input
4 1
1 2 1 5
Output
2
+ 3 2
- 4 1

Input
4 1
1 2 3 4
Output
0
*/

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