Problem: Problem

Beans are popular at Otsu University. N beans are lined up in a straight line. Each is numbered from 0 to N-1, and the hardness of the i-th bean is ai.

Cyan considers the ideal bean hardness to be D. However, Cyan doesn't want to go get the beans that are too far away because he is troublesome. Therefore, Cyan wants to know the beans whose hardness is closest to D among the l-th to r-th beans.

Cyan asks Q questions, so create a program to find the minimum value of | bean hardness − D | in the closed interval [l, r] for each question. (However, | a | represents the absolute value of a.)

Constraints

The input satisfies the following constraints.

* 1 ≤ N ≤ 105
* 0 ≤ | ai | ≤ 106 (0 ≤ i ≤ N−1)
* 1 ≤ Q ≤ 105
* 0 ≤ Di ≤ 106
* 0 ≤ li ≤ ri ≤ N-1

Input

The input is given in the following format.


N
a0 a1 ... aN−1
Q
l0 r0 D0
l1 r1 D1
..
..
..
lQ−1 rQ−1 DQ−1


The first line is given one integer N. On the second line, N integers are given, separated by blanks. On the third line, the number of queries is given as one integer Q. The query values ​​l, r, and D are given in the following 4 to 3 + Q lines.

Output

For each query, output the absolute value of the difference between D and the hardness of the bean closest to hardness D among the [l, r] th beans on one line.

Examples

Input

3
1 2 3
3
0 2 2
0 2 4
0 0 2


Output

0
1
1


Input

10
4 5 0 21 9 100 12 9 0 8
5
0 3 20
2 5 100
8 9 9
5 5 10
0 9 20


Output

1
0
1
90
1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using Int = long long;

struct FullyIndexableDictionary{
  int len,blk;
  vector<int> bit,sum;
  
  FullyIndexableDictionary(){}
  FullyIndexableDictionary(int len)
    :len(len),blk((len+31)>>5),bit(blk,0),sum(blk,0){}
  
  void set(int k){
    bit[k>>5]|=1<<(k&31);
  }

  void build(){
    sum[0]=0;
    for(int i=1;i<blk;i++)
      sum[i]=sum[i-1]+__builtin_popcount(bit[i-1]);
  }

  bool operator[](int k) const{
    return bool((bit[k>>5]>>(k&31))&1);
  }
  
  int rank(int k){
    return sum[k>>5]+__builtin_popcount(bit[k>>5]&((1<<(k&31))-1));
  }
  
  int rank(bool v,int k){
    return (v?rank(k):k-rank(k));
  }

  int select(bool v,int k){
    if(k<0||rank(v,len)<=k) return -1;
    int low=0,high=len;
    while(low+1<high){
      int mid=(low+high)>>1;
      if(rank(v,mid)>=k+1) high=mid;
      else low=mid;
    }
    return high-1;
  }

  int select(bool v,int i,int l){
    return select(v,i+rank(v,l));
  }
};

template<class T,int MAXLOG>
struct WaveletMatrix{
  int len;
  FullyIndexableDictionary mat[MAXLOG];
  int zs[MAXLOG],buff1[MAXLOG],buff2[MAXLOG];
  static const T npos=-1;
  
  int freq_dfs(int d,int l,int r,T val,T a,T b){
    if(l==r) return 0;
    if(d==MAXLOG) return (a<=val&&val<b)?r-l:0;
    T nv=1ULL<<(MAXLOG-d-1)|val;
    T nnv=((1ULL<<(MAXLOG-d-1))-1)|nv;
    if(nnv<a||b<=val) return 0;
    if(a<=val&&nnv<b) return r-l;
    int lc=mat[d].rank(1,l),rc=mat[d].rank(1,r);
    return freq_dfs(d+1,l-lc,r-rc,val,a,b)
      +freq_dfs(d+1,lc+zs[d],rc+zs[d],nv,a,b);
  }
  
  WaveletMatrix(vector<T> data){
    len=data.size();
    vector<T> l(len),r(len);
    for(int dep=0;dep<MAXLOG;dep++){
      mat[dep]=FullyIndexableDictionary(len+1);
      int p=0,q=0;
      for(int i=0;i<len;i++){
	bool k=(data[i]>>(MAXLOG-(dep+1)))&1;
	if(k) r[q++]=data[i],mat[dep].set(i);
	else  l[p++]=data[i];
      }
      zs[dep]=p;
      mat[dep].build();
      swap(l,data);
      for(int i=0;i<q;i++) data[p+i]=r[i];
    }
  }
  
  T access(int k){
    T res=0;
    bool bit;
    for(int dep=0;dep<MAXLOG;dep++){
      bit=mat[dep][k];
      res=(res<<1)|bit;
      k=mat[dep].rank(bit,k)+zs[dep]*dep;
    }
    return res;
  }

  // return the number of v in [0,k)
  int rank(T v,int k){
    int l=0,r=k;
    for(int dep=0;dep<MAXLOG;dep++){
      buff1[dep]=l;buff2[dep]=r;
      bool bit=(v>>(MAXLOG-(dep+1)))&1;
      l=mat[dep].rank(bit,l)+zs[dep]*bit;
      r=mat[dep].rank(bit,r)+zs[dep]*bit;
    }
    return r-l;
  }

  // return the position of k-th v
  int select(T v,int k){
    rank(v,len);
    for(int dep=MAXLOG-1;dep>=0;dep--){
      bool bit=(v>>(MAXLOG-(dep+1)))&1;
      k=mat[dep].select(bit,k,buff1[dep]);
      if(k>=buff2[dep]||k<0) return -1;
      k-=buff1[dep];
    }
    return k;
  }

  int select(T v,int k,int l){
    return select(v,k+rank(v,l));
  }

  // return k-th largest value in [l,r)
  T quantile(int l,int r,int k){
    if(r-l<=k||k<0) return -1;
    T res=0;
    for(int dep=0;dep<MAXLOG;dep++){
      int p=mat[dep].rank(1,l);
      int q=mat[dep].rank(1,r);
      if(q-p>k){
	l=p+zs[dep];
	r=q+zs[dep];
	res|=(1<<(MAXLOG-(dep+1)));
      }else{
	k-=(q-p);
	l-=p;
	r-=q;
      }
    }
    return res;
  }
  
  T rquantile(int l,int r,int k){
    return quantile(l,r,r-l-k-1);
  }
  
  // return number of points in [left, right) * [lower, upper)
  int rangefreq(int left,int right,T lower,T upper){
    return freq_dfs(0,left,right,0,lower,upper);
  }

  pair<int, int> ll(int l,int r,T v){
    int res=0;
    for(int dep=0;dep<MAXLOG;dep++){
      buff1[dep]=l;buff2[dep]=r;
      bool bit=(v>>(MAXLOG-(dep+1)))&1;
      if(bit) res+=r-l+mat[dep].rank(bit,l)-mat[dep].rank(bit,r);
      l=mat[dep].rank(bit,l)+zs[dep]*bit;
      r=mat[dep].rank(bit,r)+zs[dep]*bit;
    }
    return make_pair(res,r-l);    
  }
  
  int lt(int l,int r,T v){
    auto p=ll(l,r,v);
    return p.first;
  }
  
  int le(int l,int r,T v){
    auto p=ll(l,r,v);
    return p.first+p.second;
  }
  
  T succ(int l,int r,T v){
    int k=le(l,r,v);
    return k==r-l?npos:rquantile(l,r,k);
  }

  T pred(int l,int r,T v){
    int k=lt(l,r,v);
    return k?rquantile(l,r,k-1):npos;
  }
};



template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  int n;
  scanf("%d",&n);

  const int OFS = 1e6;
  vector<int> a(n);
  for(int i=0;i<n;i++) scanf("%d",&a[i]),a[i]+=OFS;
  WaveletMatrix<int, 21> wm(a);

  int q;
  scanf("%d",&q);
  for(int i=0;i<q;i++){
    int l,r,d;
    scanf("%d %d %d",&l,&r,&d);
    r++;
    d+=OFS;
    int ans=OFS*2;
    if(wm.rank(d,l)<wm.rank(d,r)) ans=0;
    int succ=wm.succ(l,r,d);
    if(~succ) chmin(ans,abs(succ-d));
    int pred=wm.pred(l,r,d);
    if(~pred) chmin(ans,abs(pred-d));
    //cout<<pred<<":"<<d<<":"<<succ<<endl;
    printf("%d\n",ans);
  }
  return 0;
}