[ Educational Codeforces Round 65 (Rated for Div. 2)][二分]

https://codeforc.es/contest/1167/problem/E

E. Range Deleting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array consisting of nn integers a1,a2,…,ana1,a2,…,an and an integer xx. It is guaranteed that for every ii, 1≤ai≤x1≤ai≤x.

Let's denote a function f(l,r)f(l,r) which erases all values such that l≤ai≤rl≤ai≤r from the array aa and returns the resulting array. For example, if a=[4,1,1,4,5,2,4,3]a=[4,1,1,4,5,2,4,3], then f(2,4)=[1,1,5]f(2,4)=[1,1,5].

Your task is to calculate the number of pairs (l,r)(l,r) such that 1≤l≤r≤x1≤l≤r≤x and f(l,r)f(l,r) is sorted in non-descending order. Note that the empty array is also considered sorted.

Input

The first line contains two integers nn and xx (1≤n,x≤1061≤n,x≤106) — the length of array aa and the upper limit for its elements, respectively.

The second line contains nn integers a1,a2,…ana1,a2,…an (1≤ai≤x1≤ai≤x).

Output

Print the number of pairs 1≤l≤r≤x1≤l≤r≤x such that f(l,r)f(l,r) is sorted in non-descending order.

Examples

input

Copy

3 3
2 3 1

output

Copy

4

input

Copy

7 4
1 3 1 2 2 4 3

output

Copy

6

Note

In the first test case correct pairs are (1,1)(1,1), (1,2)(1,2), (1,3)(1,3) and (2,3)(2,3).

In the second test case correct pairs are (1,3)(1,3), (1,4)(1,4), (2,3)(2,3), (2,4)(2,4), (3,3)(3,3) and (3,4)(3,4).

 题意：有一个含有n个元素的数组，数组元素的范围是 1<=ai<=x，你可以删除值在(l,r)的元素，问有多少种方案使删除后数组成为非严格单调递增数组

题解：固定左边界L，则右边界呈现单调性，所以可以二分。判断条件为考虑删除（L,R）之后，(1)L左边的呈现非严格单调递增，(2)R右边呈现非严格单调递增，(3)并且原数组中比L左边某一元素大并且在该元素左边的最大值应<=R，预处理一下即可二分

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define debug(x) cout<<"["<<#x<<"]"<<x<<endl;
const int maxn=1e6+5;
const int inf=1e8;
int a[maxn],b[maxn],maxx2;
bool check(ll x0,ll len){
    if(b[x0-1]>x0+len-1)return false;
    if(x0+len-1<maxx2)return false;
    return true;
}
int main() {
    int n,x;
    scanf("%d%d",&n,&x);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    int maxx=a[n];
    int minn=x;
    for(int i=n-1;i>=1;i--){
        if(maxx>=a[i]){
            maxx=a[i];
        }
        else{
            minn=min(a[i],minn);
        }
    }
    maxx=a[1];
    for(int i=2;i<=n;i++){
        if(maxx>a[i]){
            b[a[i]]=max(b[a[i]],maxx);
            maxx2=max(maxx2,a[i]);
        }
        else{
            maxx=a[i];
        }
    }
    for(int i=1;i<=x;i++){
        b[i]=max(b[i],b[i-1]);
    }
    ll aans=0;
    for(int i=1;i<=minn;i++){
        ll l=1;
        ll r=x-i+1;
        ll ans=-1;
        while(l<=r){
            ll mid=(l+r)/2;
            if(check(i,mid)){
                ans=mid;
                r=mid-1;
            }
            else{
                l=mid+1;
            }
        }
        if(ans!=-1){
            aans+=x-(i+ans-1)+1;
        }
    }
    printf("%lld
",aans);
    return 0;
}

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