Strip

B. Strip

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alexandra has a paper strip with n numbers on it. Let's call them a_i from left to right.

Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:

• Each piece should contain at least l numbers.
• The difference between the maximal and the minimal number on the piece should be at most s.

Please help Alexandra to find the minimal number of pieces meeting the condition above.

Input

The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 10⁵, 0 ≤ s ≤ 10⁹, 1 ≤ l ≤ 10⁵).

The second line contains n integers a_i separated by spaces ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

Examples

Input

7 2 2
1 3 1 2 4 1 2

Output

3

Input

7 2 2
1 100 1 100 1 100 1

Output

-1

Note

For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2].

For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists.

我先说几句 我*************

不到四十行的sbdp写了两天。。。。。。。到现在还没想懂

不说了泪奔

#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
#define N 200010
#define inf 1000000009
int n,S,l,last=1;
int dp[N],mn[3*N],a[N];
multiset<int> s;
set<int> v;
void update(int l,int r,int x,int pos,int num)
{
    if(l==r)
    {
        mn[x]=num;
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid) update(l,mid,x<<1,pos,num);
    else update(mid+1,r,x<<1|1,pos,num);
    mn[x]=min(mn[x<<1],mn[x<<1|1]);
}
int query(int l,int r,int x,int a,int b)
{
    if(l>b||r<a) return inf;
    if(l>=a&&r<=b) return mn[x];
    int mid=(l+r)>>1;
    return min(query(l,mid,x<<1,a,b),query(mid+1,r,x<<1|1,a,b));
}
int main()
{
    memset(mn,0x3f3f,sizeof(mn));
    scanf("%d%d%d",&n,&S,&l);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    /*
        dp[i]:到i位置能分多少段 
    */
    for(int i=1;i<=n;i++)
    {
        s.insert(a[i]);
        while(*s.rbegin()-*s.begin()>S) 
        {
            s.erase(s.find(a[last]));
            last++;
        }
        if(i-last+1>=l&&last==1) dp[i]=1; else 
        dp[i]=query(1,n,1,last-1,i-l)+1;
        update(1,n,1,i,dp[i]);
    }
    printf("%d\n",dp[n]>=inf?-1:dp[n]);
    return 0;
}