A

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing 2 integers l and r, please find out the biggest water source between al and ar.

Input

First you are given an integer T(T≤10) indicating the number of test cases. For each test case, there is a number n(0≤n≤1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an, and each integer is in {1,...,106}. On the next line, there is a number q(0≤q≤1000) representing the number of queries. After that, there will be q lines with two integers l and r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.

Output

For each query, output an integer representing the size of the biggest water source.

Sample Input

3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3

Sample Output

100
2
3
4
4
5
1
999999
999999
1

求区间内最大的水源，之前区间dp没法做，会T，用rmq

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define prf(x) printf("%d
",x) 
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
//#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=1e3+10;
int a[N],MAX[N][20];
void first(int n)
{
	for(int j=1;(1<<j)<=n;j++)
	{
		for(int i=1;i+(1<<j)-1<=n;i++)
		{
			MAX[i][j]=max(MAX[i][j-1],MAX[i+(1<<(j-1))][j-1]);
			//MIN[i][j]=min(MIN[i][j-1],MIN[i+(1<<(j-1))][j-1];
		}
	}
}
int solve(int l,int r)
{
	int x=0;
	while(l-1+(1<<x)<=r) x++;
	x--;
	return max(MAX[l][x],MAX[r-(1<<x)+1][x]);
}
int main()
{
	int re,n,tot,l,r;scf(re);
	while(re--)
	{
		mm(a,0);
		mm(MAX,0);
		scf(n);
		rep(i,1,n+1)
		{
			scf(a[i]);
			MAX[i][0]=a[i];
		}
		first(n);
		scf(tot);
		while(tot--)
		{
			scff(l,r);
			prf(solve(l,r));
		}
	}
	return 0;
}