CodeForces

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, ..., a_n (1 ≤ a_i ≤ 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 2
1 1 2 2
1 3 2
1 4 2

Output

1
-1

Input

5 3
1 2 1 3 2
2 5 3
1 2 3
5 5 2

Output

2
1
2

题意：给定N个数，Q个询问，每个询问给出L，R，K，求这个区间出现次数大于num=(R-L+1)/K的最小数，没有则输出-1。

思路：参照上一篇博客的“主席树求区间众数”，这一题也差不多。 同样用主席树记录前缀出现次数，线段树上跑的时候，如果这个区间的出现次数小于num，那么忽略这个区间。如果大于大于num，那么再去子区间找是否寻在答案。

（最近状态不错额，做的几个题都有手感。ORZ

#include<bits/stdc++.h>
using namespace std;
const int maxn=300010;
struct node{ 
    int val,l,r; 
    node() {} 
    node(int L,int R,int V):l(L),r(R),val(V){}
}s[maxn*20];
int rt[maxn],cnt,ans;
void insert(int &now,int pre,int L,int R,int pos,int val)
{
    s[now=++cnt]=node(s[pre].l,s[pre].r,s[pre].val+val);
    if(L==R) return ;
    int Mid=(L+R)>>1;
    if(pos<=Mid) insert(s[now].l,s[pre].l,L,Mid,pos,val);
    else insert(s[now].r,s[pre].r,Mid+1,R,pos,val);
}
int query(int now,int pre,int L,int R,int times)
{
    if(L==R) return L;
    int Mid=(L+R)>>1;
    int res=maxn,tmp;
    if(s[s[now].l].val-s[s[pre].l].val>times){
         tmp=query(s[now].l,s[pre].l,L,Mid,times);
         if(tmp!=-1) res=min(res,tmp);
    }
    if(s[s[now].r].val-s[s[pre].r].val>times){
         tmp=query(s[now].r,s[pre].r,Mid+1,R,times);
         if(tmp!=-1) res=min(res,tmp);
    }
    if(res==maxn) res=-1;
    return res;
}
int main()
{
    int N,K,Q,x,y,i,j;
    scanf("%d%d",&N,&Q);
    for(i=1;i<=N;i++){
        scanf("%d",&x);
        insert(rt[i],rt[i-1],1,N,x,1);
    }
    for(i=1;i<=Q;i++){
        scanf("%d%d%d",&x,&y,&K);
        printf("%d
",query(rt[y],rt[x-1],1,N,(y-x+1)/K));
    }
    return 0;
}