CodeForces

For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is guaranteed that the answer is not greater than 8·10¹⁸.

Input

First line contain two integer values n and k (1 ≤ n ≤ 10⁵, 0 ≤ k ≤ 10) — the length of sequence and the number of elements in increasing subsequences.

Next n lines contains one integer a_i (1 ≤ a_i ≤ n) each — elements of sequence. All values a_i are different.

Output

Print one integer — the answer to the problem.

Examples

Input

Output

题意：求多少个LIS，其长度为K+1。

思路：用dp[k][x]表示以x结尾的长度为k的LIS。那么暴力的DP复杂度是O（N^2）。

建立k棵线段树，每棵树N个叶子节点，第i棵树的第j个叶子节点表示以j为结尾的长度为i的LIS个数。那么Tree[i][j]+=Tree[i-1][1到j-1]。

(直接写线段树也可以，这里练习一下主席树。当然最好写的还是树状数组！

相比的话，主席树应该还是好写的。

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int maxn=100010;
int rt[maxn],cnt; ll ans;
struct node
{
    int l,r; ll val;
    node(){ l=r=0;val=0;}
    node(int L,int R,ll V):l(L),r(R),val(V){}
}s[maxn*30];
ll query(int Now,int L,int R,int l,int r)
{
    if(l<=L&&r>=R) return s[Now].val;
    ll res=0; int Mid=(L+R)>>1;
    if(l<=Mid) res+=query(s[Now].l,L,Mid,l,r);
    if(r>Mid) res+=query(s[Now].r,Mid+1,R,l,r);
    return res;
}
void update(int &Now,int L,int R,int pos,ll val)
{
    if(Now==0) Now=++cnt;
    s[Now].val+=val;
    if(L==R) return ; int Mid=(L+R)>>1;
    if(pos<=Mid) update(s[Now].l,L,Mid,pos,val);
    else update(s[Now].r,Mid+1,R,pos,val);
}
int main()
{
    int N,K,i,j,x;
    scanf("%d%d",&N,&K);
    for(i=1;i<=N;i++){
        scanf("%d",&x); x+=2;
        update(rt[1],1,N+2,x,1);
        for(j=0;j<=K;j++) {
            ll tmp=query(rt[j],1,N+2,1,x-1);
            update(rt[j+1],1,N+2,x,tmp);
        }
    }
    ans=query(rt[K+1],1,N+2,1,N+2);
    printf("%I64d
",ans);
    return 0;
}