Testing Round #12 C

Description

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 ai (1 ≤ a_i≤ n) each — elements of sequence. All values ai are different.

Output

Print one integer — the answer to the problem.

Examples

input

5 2
1
2
3
5
4

output

7
题意:求长度为k+1的上升子序列有多少个
解法：sum=dp[0][k+1]+dp[1][k+1]+dp[2][k+1]+....dp[n][k+1]
dp[x][y]是x为上升子序列最后一个元素，长度为y的个数
用树状数组维护，更新的是num为上升子序列最后一个元素，长度为j时，加上num为上升子序列最后一个元素，长度为j-1时个数
最后求sum(n,m+1)总和

#include <bits/stdc++.h>                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            .h>
using namespace std;
#define ll long long
ll n,m;
ll dp[200000][20];
ll bit(ll x)
{
    return x&(-x);
}
void up(ll x,ll y,ll ans)
{
    while(x<200000)
    {
        dp[x][y]+=ans;
        x+=bit(x);
    }
}
ll sum(ll x,ll y)
{
    ll ans=0;
    while(x>0)
    {
        ans+=dp[x][y];
        x-=bit(x);
    }
    return ans;
}
int main()
{
    cin>>n>>m;
    up(1,0,1);
    for(int i=1;i<=n;i++)
    {
        ll num;
        cin>>num;
        for(int j=m+1;j>=1;j--)
        {
            up(num,j,sum(num,j-1));
        }
    }
    cout<<sum(n,m+1)<<endl;
    return 0;
}