*【CodeForces - 574A】Bear and Elections （优先队列，水题模拟）

题干：

Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.

There are n candidates, including Limak. We know how many citizens are going to vote for each candidate. Now i-th candidate would get ai votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.

Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?

Input

The first line contains single integer n (2 ≤ n ≤ 100) - number of candidates.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000) - number of votes for each candidate. Limak is candidate number 1.

Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.

Output

Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.

Examples

Input

5
5 1 11 2 8

Output

Input

4
1 8 8 8

Output

Input

2
7 6

Output

Note

In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.

In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.

In the third sample Limak is a winner without bribing any citizen.

题目大意：

n个人参加选举，现在知道目前n个人各自的得票数，现在要让1号人获胜（也就是1号的得票数严格大于其他人的得票数），问最少要贿赂几个人？

解题报告：

直接优先队列可以搞出来。

AC代码：


#include<bits/stdc++.h>using namespace std;
priority_queue<int> pq;
int main()
{int n,ans=0,my,tmp;scanf("%d%d",&n,&my);//ans相当于a[1]for(int i = 2; i<=n; i++){scanf("%d",&tmp);if(tmp >= my) pq.push(tmp);}if(pq.empty()) {puts("0");return 0;}while(my<=pq.top()){my++;ans++;tmp = pq.top();pq.pop();pq.push(tmp-1);}printf("%d\n",ans);return 0;
}

wa代码：

#include<bits/stdc++.h>using namespace std;
const int MAX = 1e5 +5;
int a[MAX];
int main()
{int n,my,top=0,tmp;cin>>n;scanf("%d",&my);for(int i = 2; i<=n; i++) {scanf("%d",&tmp);if(tmp >= my) a[++top] = tmp;}if(top == 0) {puts("0");return 0;}int sum = my;for(int i = 1; i<=top; i++) {sum += a[i];}double ans = ceil(sum*1.0/(top+1));if(sum%(top+1) == 0) {printf("%d\n",(int)ans-my + 1);}else printf("%d\n",(int)ans-my);return 0 ;}

总结：

刚开始想错了，想直接找到平均值然后看是否需要加1就可以了。但是这样是不对的，因为比如1,897,2这三个数，平均值是300，但是你答案不能直接在300附近找，因为这样看的话相当于897也分给了2一部分，才能平均成300，但是依据题意，只有1号选手可以贿赂其他人，而对于其他人，得票数只能减少不能增加的！（也就是，其他人都不能贿赂别人）。所以这种策略（直接算平均值）是失效的，对于这个题。

emmm想一下，这题也可以二分答案去做吗？

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*【CodeForces - 574A】Bear and Elections （优先队列，水题模拟）

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