【Educational Codeforces Round 33】 D. Credit Card (贪心)

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本文介绍了一个关于信用卡收支管理的问题及解决方法。主要内容包括如何确保信用卡余额在规定限制内，并在特定检查日保持非负数。通过算法实现了最小次数的存款操作以满足条件。

Recenlty Luba got a credit card and started to use it. Let's consider n consecutive days Luba uses the card.

She starts with 0 money on her account.

In the evening of i-th day a transaction ai occurs. If ai > 0, then ai bourles are deposited to Luba's account. If ai < 0, then ai bourles are withdrawn. And if ai = 0, then the amount of money on Luba's account is checked.

In the morning of any of n days Luba can go to the bank and deposit any positive integer amount of burles to her account. But there is a limitation: the amount of money on the account can never exceed d.

It can happen that the amount of money goes greater than d by some transaction in the evening. In this case answer will be «-1».

Luba must not exceed this limit, and also she wants that every day her account is checked (the days when ai = 0) the amount of money on her account is non-negative. It takes a lot of time to go to the bank, so Luba wants to know the minimum number of days she needs to deposit some money to her account (if it is possible to meet all the requirements). Help her!

Input

The first line contains two integers n, d (1 ≤ n ≤ 105, 1 ≤ d ≤ 109) —the number of days and the money limitation.

The second line contains n integer numbers a1, a2, ... an ( - 104 ≤ ai ≤ 104), where ai represents the transaction in i-th day.

Output

Print -1 if Luba cannot deposit the money to her account in such a way that the requirements are met. Otherwise print the minimum number of days Luba has to deposit money.

Examples

input
5 10
-1 5 0 -5 3

output
0

input
3 4
-10 0 20

output
-1

input
5 10
-5 0 10 -11 0

output
2

#include <bits/stdc++.h>
using namespace std;
int a[100001], pre[100001], maxsuf[100001];
int main(){
	int n, d;
	scanf("%d %d", &n, &d);
	for(int i = 1; i <= n; ++i){
		scanf("%d", &a[i]);
	}
	pre[0] = 0;
	for(int i = 1; i <= n; ++i){
		pre[i] = pre[i - 1] + a[i];
	}
	maxsuf[n] = pre[n];
	for(int i = n - 1; i >= 1; --i){
		maxsuf[i] = max(maxsuf[i + 1], pre[i]);
	}
	long long add = 0;
	int ans = 0;
	for(int i = 1; i <= n; ++i){
		if(a[i] == 0){
			if(add + pre[i] < 0){
				add += d - (maxsuf[i] + add);
				ans++;
			}
			if(add + pre[i] < 0){
				printf("-1\n");
				return 0;
			}
		}
		else{
			if(add + pre[i] > d){
				printf("-1\n");
				return 0;
			}
		}
	}
	printf("%d\n", ans);
}

/*
题意：
一张信用卡，白天主人可以去银行存钱，随意存多少钱，但必须是正整数，且保证账户金额
不超过d，晚上的时候会对信用卡金额做一些变化，如果是a[i]是0则保证此时信用卡金额
必须大于等于0。问主人最少需要去充多少钱，如果出现违反规则的情况输出-1。

思路：
首先由于每次充钱我们只需要保证账户金额不超过d就可以无限充钱，那么我们不会因为检查
时金额为负数而输出-1，因为我们一定有能力在检查的那天白天把金额充值到正数。现在的
问题是，任何一天的白天金额不能超过d，且希望充值次数最少。所以在每次必须充值的时候，
我们尽量多充一些，充多少由后面的前缀最大值定。假设未来的最大前缀是pre[j]，那么此时
最多只能充d - pre[j]。
*/