Codeforces Round #429 (Div. 1) D. Destiny(主席树)

D. Destiny

time limit per test

2.5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Once, Leha found in the left pocket an array consisting of n integers, and in the right pocket q queries of the form l r k. If there are queries, then they must be answered. Answer for the query is minimal x such that x occurs in the interval l r strictly more than times or  - 1 if there is no such number. Help Leha with such a difficult task.

Input

First line of input data contains two integers n and q (1 ≤ n, q ≤ 3·105) — number of elements in the array and number of queries respectively.

Next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — Leha's array.

Each of next q lines contains three integers l, r and k (1 ≤ l ≤ r ≤ n, 2 ≤ k ≤ 5) — description of the queries.

Output

Output answer for each query in new line.

Examples

input

4 21 1 2 21 3 21 4 2

output

1-1

input

5 31 2 1 3 22 5 31 2 35 5 2

output

212

分析:k很小,只有2到5,然后我们考虑将l 到 r的数排序后,设 p = ,那么答案一定是第k*p个数,因为如果答案出现多于p次那么一定会跨越一个边界,因为k很小所以我们顶多试5次,直接用主席树求解就可以了.

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define N 300005using namespace std;struct Thing{int ls,rs,sum = 0;}tr[N*30];int n,q,p,cnt,a[N],rt[N];bool flag;void Build(int &node,int l,int r){node = ++cnt;tr[node].sum = 0;if(l == r) return;int mid = (l+r)>>1;Build(tr[node].ls,l,mid);Build(tr[node].rs,mid+1,r);}void Insert(int pre,int &node,int l,int r,int pos){node = ++cnt;tr[node] = tr[pre];tr[node].sum++;if(l == r) return;int mid = (l+r)>>1;if(pos <= mid) Insert(tr[pre].ls,tr[node].ls,l,mid,pos);else Insert(tr[pre].rs,tr[node].rs,mid+1,r,pos);}int Query(int pre,int node,int l,int r,int k){    if(l == r)     {flag = (tr[node].sum - tr[pre].sum) > p; return l;}    int cmp = tr[tr[node].ls].sum - tr[tr[pre].ls].sum,mid = (l+r)>>1;    if(cmp >= k)return Query(tr[pre].ls,tr[node].ls,l,mid,k);    else return Query(tr[pre].rs,tr[node].rs,mid+1,r,k-cmp);}int main(){scanf("%d%d",&n,&q);for(int i = 1;i <= n;i++) scanf("%d",&a[i]);Build(rt[0],1,n);for(int i = 1;i <= n;i++) Insert(rt[i-1],rt[i],1,n,a[i]); for(int i = 1;i <= q;i++){int l,r,k,ans;scanf("%d%d%d",&l,&r,&k);p = (r - l + 1)/k,ans = -1;if(!p){cout<<Query(rt[l-1],rt[r],1,n,1)<<endl;continue;}flag = false;for(int j = p;j <= (r - l + 1);j += p){int t = Query(rt[l-1],rt[r],1,n,j);if(flag){ans = t;break;}}cout<<ans<<endl;}}