Goods transportation

Goods transportation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n cities located along the one-way road. Cities are numbered from 1 to n in the direction of the road.

The i-th city had produced pi units of goods. No more than si units of goods can be sold in the i-th city.

For each pair of cities i and j such that 1 ≤ i < j ≤ n you can no more than once transport no more than c units of goods from the city i to the city j. Note that goods can only be transported from a city with a lesser index to the city with a larger index. You can transport goods between cities in any order.

Determine the maximum number of produced goods that can be sold in total in all the cities after a sequence of transportations.

Input

The first line of the input contains two integers n and c (1 ≤ n ≤ 10 000, 0 ≤ c ≤ 109) — the number of cities and the maximum amount of goods for a single transportation.

The second line contains n integers pi (0 ≤ pi ≤ 109) — the number of units of goods that were produced in each city.

The third line of input contains n integers si (0 ≤ si ≤ 109) — the number of units of goods that can be sold in each city.

Output

Print the maximum total number of produced goods that can be sold in all cities after a sequence of transportations.

Examples

input

3 0
1 2 3
3 2 1

output

4

input

5 1
7 4 2 1 0
1 2 3 4 5

output

12

input

4 3
13 10 7 4
4 7 10 13

output

34
分析：考虑最大流等于最小割，从小到大dp；
　　　dp[i][j]表示前i个点有j个点在最小割点集里，
　　　则dp[i][j]=min(dp[i-1][j-1]+s[i],dp[i-1][j]+j*c+p[i]);
　　　dp[i-1][j-1]+s[i]表示i留在s-割的代价，dp[i-1][j]+j*c+p[i]表示i留在t-割，除了要花费p[i]外，还要花费j*c的代价；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<ll,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=1e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t;
ll p[maxn],s[maxn],ans[maxn],c;
int main()
{
    int i,j;
    scanf("%d%lld",&n,&c);
    rep(i,1,n)scanf("%lld",&p[i]);
    rep(i,1,n)scanf("%lld",&s[i]);
    rep(i,1,n)
    {
        ans[i]=1e18;
        for(j=i;j>=1;j--)
        {
            ans[j]=min(ans[j]+j*c+p[i],ans[j-1]+s[i]);
        }
        ans[0]+=p[i];
    }
    ll ret=1e18;
    rep(i,0,n)ret=min(ret,ans[i]);
    printf("%lld
",ret);
    //system("Pause");
    return 0;
}