2020-2021 ICPC, NERC, Southern and Volga Russian Regional Contest B. Bakery

题目链接
Monocarp would like to open a bakery in his local area. But, at first, he should figure out whether he can compete with other shops.

Monocarp plans that the bakery will work for nnn days. On the iii-th day, aia_iai loaves of bread will be baked in the morning before the opening. At the end of the nnn-th day, Monocarp will sell all the remaining bread that wasn’t sold earlier with a huge discount.

Because of how bread is stored, the bakery seller sells the bread in the following order: firstly, he sells the loaves that were baked that morning; secondly, he sells the loaves that were baked the day before and weren’t sold yet; then the loaves that were baked two days before and weren’t sold yet, and so on. That’s why some customers may buy a rather stale bread and will definitely spread negative rumors.

Let’s define loaf spoilage as the difference between the day it was baked and the day it was sold. Then the unattractiveness of the bakery will be equal to the maximum spoilage among all loaves of bread baked at the bakery.

Suppose Monocarp’s local area has consumer demand equal to kkk, it means that each day kkk customers will come to the bakery and each of them will ask for one loaf of bread (the loaves are sold according to the aforementioned order). If there is no bread left, then the person just doesn’t buy anything. During the last day sale, all the remaining loaves will be sold (and they will still count in the calculation of the unattractiveness).

Monocarp analyzed his competitors’ data and came up with mmm possible consumer demand values k1,k2,…,kmk_1, k_2, \dots, k_mk1,k2,…,km, and now he’d like to calculate the unattractiveness of the bakery for each value of demand. Can you help him?

Input

The first line contains two integers nnn and mmm (1≤n≤2⋅1051 \le n \le 2 \cdot 10^51≤n≤2⋅105; 1≤m≤2⋅1051 \le m \le 2 \cdot 10^51≤m≤2⋅105) — the number of days the bakery is open and the number of possible values of consumer demand.

The second line contains nnn integers a1,a2,…,ana_1, a_2, \dots, a_na1,a2,…,an (1≤ai≤1091 \le a_i \le 10^91≤ai≤109) — the number of bread loaves that will be baked each day.

The third line contains mmm integers k1,k2,…,kmk_1, k_2, \dots, k_mk1,k2,…,km (1≤k1<k2<⋯<km≤1091 \le k_1 < k_2 < \dots < k_m \le 10^91≤k1<k2<⋯<km≤109) — the possible consumer demand values in the ascending order.

Output

Print mmm integers: for each consumer demand, print the unattractiveness of the bakery.

Examples

input

5 4
5 2 1 3 7
1 3 4 10

output

4 2 1 0

input

8 9
3 1 4 1 5 9 2 6
1 2 3 4 5 6 7 8 9

output

7 5 3 3 2 1 1 1 0

Note

In the first example, let’s describe what happens for couple consumer demands:

If consumer demand is equal to 111:

at day 111: 555 loaves are baked and only 111 is sold with spoilage equal to 1−1=01 - 1 = 01−1=0;
at day 222: 444 loaves are left and 222 more are baked. Only 111 loaf was sold and it was the loaf baked today with spoilage 2−2=02 - 2 = 02−2=0;
at day 333: 444 loaves from the first day and 111 loaf from the second day left. One more loaf was baked and was sold this day with spoilage 3−3=03 - 3 = 03−3=0;
at day 444: 444 loaves from the first day and 111 loaf from the second day left. 333 more loaves were baked and one of them was sold this day with spoilage 4−4=04 - 4 = 04−4=0;
at day 555: 444 loaves from the first day, 111 loaf from the second day and 222 loaves from the fourth day left. 777 more loaves were baked and, since it’s the last day, all 141414 loaves were sold. 444 loaves from the first day have the maximum spoilage equal to 5−1=45 - 1 = 45−1=4.

In total, the unattractiveness of the bakery will be equal to 444.

If consumer demand is equal to 101010 then all baked bread will be sold in the day it was baked and will have spoilage equal to 000.

首先将 aaa 求前缀和，然后在前缀和用单调栈预处理以每一天为结束的平均产量最高的连续天的值并按升序排序。
考虑维护一个区间集合，对于集合中的某个区间 [l,r][l,r][l,r] ，其左右端点 l,rl,rl,r 为右端点的任意区间内的平均值一定小于等于当前每天的平均花费 kkk ，而以区间 (l,r)(l,r)(l,r) 中任意一点为右端点一定能找到一个区间使得平均值大于 kkk ，因此区间 (l,r)(l,r)(l,r) 中任意一天面包一定有剩余，而区间端点处面包一定没有剩余。进一步可以得出第 l+1l+1l+1 天一定有当天面包的剩余，因此第 rrr 天一定用到了第 l+1l+1l+1 天的面包，且一定不会用到第 l+1l+1l+1 天之前的面包。因此 unattractiveness 为最长区间长度减 222。
从小到大遍历 kkk ，并维护区间即可得到答案。

#include<bits/stdc++.h>#define ll long long
#define ce(i, r) i==r?'\n':' '
#define fi first
#define se second
using namespace std;inline int qr() {int f = 0, fu = 1;char c = getchar();while (c < '0' || c > '9') {if (c == '-')fu = -1;c = getchar();}while (c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15);c = getchar();}return f * fu;
}const int N = 2e5 + 10;
pair<int, int> seq[N];
pair<int, ll> stk[N];
int n, m;
ll a[N];
set<int> st;
multiset<int> xst;inline void ins(int x) {auto it = st.lower_bound(x);int r = *it, l = *--it;xst.erase(xst.find(r - l));xst.insert(r - x), xst.insert(x - l);st.insert(x);
}int main() {n = qr(), m = qr();for (int i = 1; i <= n; i++)a[i] = a[i - 1] + qr();for (int i = 1, t = 1; i <= n - 1; i++) {while (t >= 2 && (__int128) (a[i] - stk[t].se) * (i - stk[t - 1].fi) <=(__int128) (a[i] - stk[t - 1].se) * (i - stk[t].fi))t--;seq[i] = {(a[i] - stk[t].se - 1) / (i - stk[t].fi) + 1, i}, stk[++t] = {i, a[i]};}seq[n] = {INT_MAX, n};sort(seq + 1, seq + n);st.insert(0), st.insert(n), xst.insert(n);for (int i = 1, now = 1; i <= m; i++) {int v = qr();while (v >= seq[now].fi)ins(seq[now++].se);printf("%d%c", *--xst.end() - 1, ce(i, m));}return 0;
}

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