Artsem is on vacation and wants to buy souvenirs for his two
teammates. There are n souvenir shops along the street. In i-th shop
Artsem can buy one souvenir for ai dollars, and he cannot buy more
than one souvenir in one shop. He doesn’t want to introduce envy in
his team, so he wants to buy two souvenirs with least possible
difference in price.

Artsem has visited the shopping street m times. For some strange
reason on the i-th day only shops with numbers from li to ri were
operating (weird? yes it is, but have you ever tried to come up with a
reasonable legend for a range query problem?). For each visit, Artsem
wants to know the minimum possible difference in prices of two
different souvenirs he can buy in the opened shops.

In other words, for each Artsem’s visit you should find the minimum
possible value of |as - at| where li ≤ s, t ≤ ri, s ≠ t. Input

The first line contains an integer n (2 ≤ n ≤ 105).

The second line contains n space-separated integers a1, …, an
(0 ≤ ai ≤ 109).

The third line contains the number of queries m (1 ≤ m ≤ 3·105).

Next m lines describe the queries. i-th of these lines contains two
space-separated integers li and ri denoting the range of shops working
on i-th day (1 ≤ li < ri ≤ n). Output

Print the answer to each query in a separate line.

解法二（線段樹套平衡樹）見這裡。
以下對於i<j，只考慮ai>aj的情況。另外的情況只需要把原序列倒過來再做一遍。
把詢問離線，從左到右加入右端點，維護每個左端點的答案，就可以知道以當前點為右端點的所有詢問的答案。
當加入ai時，首先，原來有的答案還成立，但是可以新加入一些有ai參與的答案。首先找到最大的j<i使得aj≥ai，顯然j+1往後的答案不會被修改。可以用aj−ai的值更新1..j的答案。然後再找到最大的j′<j使得ai≤aj′<aj，顯然j′+1往後的答案不會被修改，可以用aj′−ai的值更新1..j′的答案。重複這樣的操作即可，可以用線段樹維護答案。
容易看出這樣的做法最多要操作O(n)次，但是可以發現只有滿足aj′−ai<aj−aj′的j′才是值得更新的，因為任何包含j′和i的區間一定包含j′和j，這樣的答案肯定會在反向時被考慮。於是aj′的範圍縮小到了[ai,(ai+aj)/2)。換句話說，aj−ai每次至少縮小一半。這樣只用操作O(logai)次了。
但是如果每次暴力尋找j′仍是O(n)的，不可取。可以離散化以後建立一棵權值線段樹維護最後出現的位置，每掃描過一個元素就加入線段樹，不難求出某一段權值的最後出現位置。當然也可以用平衡樹來維護。
最後的複雜度是O(nlognlogai+mlogn)。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<vector>
using namespace std;
#define MP(a,b) make_pair(a,b)
const int oo=0x3f3f3f3f;
int a[100010],tag[400010],
late[400010],ord[100010],newa[100010],
ans[300010],ql[300010],qr[300010],
n,nn,m;
vector<pair<int,int> > query[300010];
int findmx(int p,int L,int R,int l,int r)
{
    if (l<=ord[L]&&ord[R]<r) return late[p];
    int mid=(L+R)/2,ret=-1;
    if (l<=ord[mid]) ret=max(ret,findmx(p*2,L,mid,l,r));
    if (ord[mid+1]<r) ret=max(ret,findmx(p*2+1,mid+1,R,l,r));
    return ret;
}
void add(int p,int L,int R,int k,int x)
{
    if (L==R)
    {
        late[p]=x;
        return;
    }
    int mid=(L+R)/2;
    if (k<=mid) add(p*2,L,mid,k,x);
    else add(p*2+1,mid+1,R,k,x);
    late[p]=max(late[p*2],late[p*2+1]);
}
void modi(int p,int L,int R,int l,int r,int x)
{
    if (l<=L&&R<=r)
    {
        tag[p]=min(tag[p],x);
        return;
    }
    int mid=(L+R)/2;
    if (l<=mid) modi(p*2,L,mid,l,r,x);
    if (r>=mid+1) modi(p*2+1,mid+1,R,l,r,x);
}
int qry(int p,int L,int R,int k)
{
    if (L==R) return tag[p];
    int mid=(L+R)/2;
    if (k<=mid)
    {
        tag[p*2]=min(tag[p*2],tag[p]);
        return qry(p*2,L,mid,k);
    }
    else
    {
        tag[p*2+1]=min(tag[p*2+1],tag[p]);
        return qry(p*2+1,mid+1,R,k);
    }
}
void solve()
{
    int x,y,p;
    memset(tag,0x3f,sizeof(tag));
    memset(late,0xff,sizeof(late));
    for (int i=1;i<=n;i++)
    {
        x=a[i];
        y=2*oo-x;
        while (x<y&&(p=findmx(1,1,nn,x,(x+y+1)/2))!=-1)
        {
            modi(1,1,n,1,p,a[p]-a[i]);
            y=a[p];
        }
        for (int j=0;j<query[i].size();j++)
            ans[query[i][j].second]=min(ans[query[i][j].second],qry(1,1,n,query[i][j].first));
        add(1,1,nn,newa[i],i);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]),ord[i]=a[i];
    sort(ord+1,ord+n+1);
    nn=unique(ord+1,ord+n+1)-ord-1;
    for (int i=1;i<=n;i++)
        newa[i]=lower_bound(ord+1,ord+nn+1,a[i])-ord;
    scanf("%d",&m);
    for (int i=1;i<=m;i++) scanf("%d%d",&ql[i],&qr[i]);
    for (int i=1;i<=m;i++) query[qr[i]].push_back(MP(ql[i],i));
    memset(ans,0x3f,sizeof(ans));
    solve();
    for (int i=1;i*2<=n;i++) swap(a[i],a[n-i+1]),swap(newa[i],newa[n-i+1]);
    memset(query,0,sizeof(query));
    for (int i=1;i<=m;i++) query[n-ql[i]+1].push_back(MP(n-qr[i]+1,i));
    solve();
    //for (int i=1;i<=m;i++) if (ans[i]==oo) printf("no:%d\n",i);
    for (int i=1;i<=m;i++) printf("%d\n",ans[i]);
}