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[][];
 ll bit(ll x)
 {
     return x&(-x);
 }
 void up(ll x,ll y,ll ans)
 {
     while(x<)
     {
         dp[x][y]+=ans;
         x+=bit(x);
     }
 }
 ll sum(ll x,ll y)
 {
     ll ans=;
     while(x>)
     {
         ans+=dp[x][y];
         x-=bit(x);
     }
     return ans;
 }
 int main()
 {
     cin>>n>>m;
     up(,,);
     for(int i=;i<=n;i++)
     {
         ll num;
         cin>>num;
         for(int j=m+;j>=;j--)
         {
             up(num,j,sum(num,j-));
         }
     }
     cout<<sum(n,m+)<<endl;
     return ;
 }