Codeforces 522D Closest Equals

You are given sequence a1, a2, ..., an and m queries lj, rj (1 ≤ lj ≤ rj ≤ n). For each query you need to print the minimum distance between such pair of elements ax and ay (x ≠ y), that:

both indexes of the elements lie within range [lj, rj], that is, lj ≤ x, y ≤ rj;
the values of the elements are equal, that is ax = ay.

The text above understands distance as |x - y|.

Input

The first line of the input contains a pair of integers n, m (1 ≤ n, m ≤ 5·105) — the length of the sequence and the number of queries, correspondingly.

The second line contains the sequence of integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109).

Next m lines contain the queries, one per line. Each query is given by a pair of numbers lj, rj (1 ≤ lj ≤ rj ≤ n) — the indexes of the query range limits.

Output

Print m integers — the answers to each query. If there is no valid match for some query, please print -1 as an answer to this query.

Examples

input

output

1
-1
2

input

6 5
1 2 1 3 2 3
4 6
1 3
2 5
2 4
1 6

output

2
2
3
-1
2

题目链接
 
题意 
　　给你n个数，m个查询，每次查询区间[l,r]中两个相同元素最近的距离的最小值。
分析 
1.设pre[i]为a[i]==a[pre[i]]且pre[i]是在i之前的距离最近的位置。那么问题就变成查询区间中最小的i-pre[i]的值，也就是区间最小值，可以用线段树处理。
2.但是，当某个点不在此区间内时，我们是不考虑的。由于这题是没有修改的，可以想到离线处理。将查询区间按左端点排序，假设当前查询的区间到了[l,r]，
那么在下标l之前的所以元素都不能考虑入内，所以我们需要消除它们对其后的最近点的影响，即使得pre[i]==x的那个pre[i]置为inf。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
using namespace std;
typedef long long LL;
const int maxn = 5e5+5;
const int mod = 772002+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1
struct node{
    int l,r,pos;
}p[maxn];

int tree[maxn*8],a[maxn],pre[maxn],ans[maxn],data[maxn];

int cmp(node a,node b){ return a.l<b.l; }

void pushup(int rt){
    tree[rt]=min(tree[rt<<1],tree[rt<<1|1]);
}
//L,R为查询区间
int query(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        return tree[rt];
    }
    int m=(r+l)>>1;
    int res=inf;
    if(L<=m){
        res=min(query(L,R,lson),res);
    }
    if(m<R){
        res=min(query(L,R,rson),res);
    }
    return res;
}

void build(int l,int r,int rt){
    if(l==r){
        tree[rt]=inf;
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int pos,int da,int l,int r,int rt){
    if(l==r){
        tree[rt]=da;
        return;
    }
    int m=(l+r)>>1;
    if(pos<=m) update(pos,da,lson);
    else update(pos,da,rson);
    pushup(rt);
}

int main(){
    int n,q;

    scanf("%d%d",&n,&q);
//    build(1,n,1);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    ms(data,-1);
    ms(tree,inf);
    map<int,int> ma; //ma[i]存储i所在的位置
    for(int i=1;i<=n;i++){
        if(i==1) pre[i]=inf;
        else{
            if(ma[a[i]]==0){
                pre[i]=inf;
            }else{
                pre[i]=ma[a[i]];
                data[ma[a[i]]]=i; //data[i]表示i位置处的数后面最近的数的下标，用作离线维护
            }
        }
        ma[a[i]]=i;
        if(pre[i]==inf) update(i,pre[i],1,n,1);
        else update(i,i-pre[i],1,n,1);
    }

    for(int i=0;i<q;i++){
        scanf("%d%d",&p[i].l,&p[i].r);
        p[i].pos=i;
    }
    sort(p,p+q,cmp);
//    for(int i=1;i<=n;i++) cout<<data[i]<<" ";
//    puts("");
    int now=1;
    for(int i=0;i<q;i++){
        if(p[i].l>now){
            for(int j=now;j<p[i].l;j++){
                update(data[j],inf,1,n,1);
                now++;
            }
        }
        ans[p[i].pos]=query(1,p[i].r,1,n,1);
        if(ans[p[i].pos]==inf) ans[p[i].pos]=-1;
    }
    for(int i=0;i<q;i++) cout<<ans[i]<<endl;
    return 0;
}