C. Increasing by Modulo

耗尽温柔 提交于 2020-01-25 03:40:43

链接:https://codeforces.com/contest/1169/problem/C

Toad Zitz has an array of integers, each integer is between 00 and m−1m−1 inclusive. The integers are a1,a2,…,ana1,a2,…,an.

In one operation Zitz can choose an integer kk and kk indices i1,i2,…,iki1,i2,…,ik such that 1≤i1<i2<…<ik≤n1≤i1<i2<…<ik≤n. He should then change aijaij to ((aij+1)modm)((aij+1)modm) for each chosen integer ijij. The integer mm is fixed for all operations and indices.

Here xmodyxmody denotes the remainder of the division of xx by yy.

Zitz wants to make his array non-decreasing with the minimum number of such operations. Find this minimum number of operations.

Input

The first line contains two integers nn and mm (1≤n,m≤3000001≤n,m≤300000) — the number of integers in the array and the parameter mm.

The next line contains nn space-separated integers a1,a2,…,ana1,a2,…,an (0≤ai<m0≤ai<m) — the given array.

Output

Output one integer: the minimum number of described operations Zitz needs to make his array non-decreasing. If no operations required, print 00.

It is easy to see that with enough operations Zitz can always make his array non-decreasing.

Examples

input

Copy

5 3
0 0 0 1 2

output

Copy

0

input

Copy

5 7
0 6 1 3 2

output

Copy

1

Note

In the first example, the array is already non-decreasing, so the answer is 00.

In the second example, you can choose k=2k=2, i1=2i1=2, i2=5i2=5, the array becomes [0,0,1,3,3][0,0,1,3,3]. It is non-decreasing, so the answer is 11.

代码:

#include <bits/stdc++.h>
using namespace std;
long long t,n,m,k,x,y,s=0,sum=0,min1=0,max1=0;
long long a[300005],b[300005];
int main()
{
    cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	int l=0,r=m;
	while(l<r)
	{
		int mid=(l+r)/2,lt=0,i;
		for(i=0;i<n;i++)
		{
			if(a[i]<=lt&&a[i]+mid>=lt|| a[i]+mid-m>=lt)
			{
				continue;
			}
			if(a[i]<lt)
			{
				break;
			}
			lt=a[i];
		}
		if(i==n)
		{
			r=mid;
		}
		else
		{
			l=mid+1;
		}
	}
	cout<<l<<endl;
	return 0;
}

 

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