B. Video Posts

Polycarp took $n$ videos, the duration of the $i$-th video is $a_i$ seconds. The videos are listed in the chronological order, i.e. the $1$-st video is the earliest, the $2$-nd video is the next, …, the $n$-th video is the last.

Now Polycarp wants to publish exactly $k$ ($1\le k\le n$) posts in Instabram. Each video should be a part of a single post. The posts should preserve the chronological order, it means that the first post should contain one or more of the earliest videos, the second post should contain a block (one or more videos) going next and so on. In other words, if the number of videos in the $j$-th post is $s_j$ then:

• $s_1+s_2+\cdots +s_k=n$ ($s_i>0$),
• the first post contains the videos: $1,2,\dots ,s_1$;
• the second post contains the videos: $s_1+1,s_1+2,\dots ,s_1+s_2$;
• the third post contains the videos: $s_1+s_2+1,s_1+s_2+2,\dots ,s_1+s_2+s_3$;
• …
• the $k$-th post contains videos: $n-s_k+1,n-s_k+2,\dots ,n$.

Polycarp is a perfectionist, he wants the total duration of videos in each post to be the same.

Help Polycarp to find such positive integer values $s_1,s_2,\dots ,s_k$ that satisfy all the conditions above.

Input

The first line contains two integers $n$ and $k$ ($1\le k\le n\le 10^5$). The next line contains $n$ positive integer numbers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 10^4$), where $a_i$ is the duration of the $i$-th video.

Output

If solution exists, print “Yes” in the first line. Print $k$ positive integers $s_1,s_2,\dots ,s_k$ ($s_1+s_2+\cdots +s_k=n$) in the second line. The total duration of videos in each post should be the same. It can be easily proven that the answer is unique (if it exists).

If there is no solution, print a single line “No”.

Example

Input

6 3
3 3 1 4 1 6

Output

Yes
2 3 1 

Input

3 1
1 10 100

Output

No

code

#include<bits/stdc++.h>
#define int long long
#define endl '\n'using namespace std;const int N = 2e5+10,INF=0x3f3f3f3f,mod=1e9+7;typedef pair<int,int> PII;int T=1;void solve(){int n,k;cin>>n>>k;int sum=0;vector<int> a(n+5,0);vector<int> ans(n+5,0);for(int i=1;i<=n;i++){cin>>a[i];sum+=a[i];}if(sum%k!=0){cout<<"No"<<endl;return;}int x = sum/k;int cnt=0;int t=0;int u=0;for(int i=1;i<=n;i++){if(t+a[i]<x){cnt++;t+=a[i];}else if(t+a[i]==x){cnt++;ans[u++]=cnt;cnt=0;t=0;}else if(t+a[i]>x){cout<<"No"<<endl;return;}}cout<<"Yes"<<endl;for(int i=0;i<u;i++) cout<<ans[i]<<" ";
}signed main(){
//	cin>>T; while(T--){solve();}return 0;
}