Balanced Lineup

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.  Lines 2.. N

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

題目大意：n頭牛給出牛的高度，查詢m次，查詢區間的最大差值（即區間內最高牛與最矮牛的差值）（線段樹求區間最大差值）

程式碼：  

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 200005
#define inf 0x3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int maxx[N<<2],minn[N<<2];
void pushup(int rt)
{
    maxx[rt]=max(maxx[rt<<1],maxx[rt<<1|1]);
    minn[rt]=min(minn[rt<<1],minn[rt<<1|1]);
}
void built(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&maxx[rt]);
        minn[rt]=maxx[rt];
        return ;
    }
    int m=(l+r)>>1;
    built(lson);
    built(rson);
    pushup(rt);
}
int querymax(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return maxx[rt];
    int ans=-1;
    int m=(l+r)>>1;
    if(L<=m) ans=max(ans,querymax(L,R,lson));
    if(m<R) ans=max(ans,querymax(L,R,rson));
    return ans;
}
int querymin(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return minn[rt];
    int ans=inf;
    int m=(l+r)>>1;
    if(L<=m) ans=min(ans,querymin(L,R,lson));
    if(m<R) ans=min(ans,querymin(L,R,rson));
    return ans;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    built(1,n,1);
    while(m--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",querymax(a,b,1,n,1)-querymin(a,b,1,n,1));
    }
    return 0;
}