There are n cities in the country where the Old Peykan lives. These cities are located on a straight line, we‘ll denote them from left to right as c₁, c₂, ..., c_n. The Old Peykan wants to travel from city c1 to cn using roads. There are (n - 1) one way roads, the i-th road goes from city c_i to city c_i + 1 and is d_i kilometers long.

The Old Peykan travels 1 kilometer in 1 hour and consumes 1 liter of fuel during this time.

Each city c_i (except for the last city c_n) has a supply of s_i liters of fuel which immediately transfers to the Old Peykan if it passes the city or stays in it. This supply refreshes instantly k hours after it transfers. The Old Peykan can stay in a city for a while and fill its fuel tank many times.

Initially (at time zero) the Old Peykan is at city c₁ and s₁ liters of fuel is transferred to it‘s empty tank from c₁‘s supply. The Old Peykan‘s fuel tank capacity is unlimited. Old Peykan can not continue its travel if its tank is emptied strictly between two cities.

Find the minimum time the Old Peykan needs to reach city c_n.

The first line of the input contains two space-separated integers m and k (1 ≤ m, k ≤ 1000). The value m specifies the number of roads between cities which is equal to n - 1.

The next line contains m space-separated integers d₁, d₂, ..., d_m (1 ≤ d_i ≤ 1000) and the following line contains m space-separated integers s₁, s₂, ..., s_m (1 ≤ s_i ≤ 1000).

4 61 2 5 22 3 3 4

10

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <stack>#include <vector>#include <queue>#include <cstring>#include <string>#include <map>#include <set>using namespace std;const int BufferSize = 1 << 16;char buffer[BufferSize], *Head, *Tail;inline char Getchar() {    if(Head == Tail) {        int l = fread(buffer, 1, BufferSize, stdin);        Tail = (Head = buffer) + l;    }    return *Head++;}int read() {    int x = 0, f = 1; char c = getchar();    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }    return x * f;}#define maxn 1010#define LL long longint n, k, d[maxn], s[maxn];int main() {	n = read(); k = read();	for(int i = 1; i <= n; i++) d[i] = read();	for(int i = 1; i <= n; i++) s[i] = read();		int ni = 1, mx = 0;	LL nt = 0, ns = 0;	for(; ni <= n + 1;) {		mx = max(mx, s[ni]);		ns += s[ni];		if(ns >= d[ni]) ;		else {			int x = (d[ni] - ns) / mx;			if(mx * x == (d[ni] - ns)) ; else x++;			nt += (LL)x * k;			ns += (LL)x * mx;		}		ns -= d[ni]; nt += d[ni]; ni++;	}		printf("%I64d\n", nt);		return 0;}