Description

You are given a sequence of n integers a₁ , a₂ , ... , a_n in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers a_i

Input

The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i

The last test case is followed by a line containing a single 0.

Output

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

Sample Input

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

Sample Output

1
4
3

題意：

題解：

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
const int MAX=100005;
int dp[MAX][20];
int mm[MAX];
int f[MAX];
void initrmq(int n,int b[])
{
    mm[0]=-1;
    for(int i=1;i<=n;i++)
    {
        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
        dp[i][0]=f[i];
    }
    for(int j=1;j<=mm[n];j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq(int x,int y)
{
    int k=mm[y-x+1];
    return max(dp[x][k],dp[y-(1<<k)+1][k]);
}
int main()
{
    int n,q,i;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&q);
        int b[MAX];
        for(i=1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            if(i==1)
            {
                f[i]=1;
                continue;
            }
            if(b[i]==b[i-1])
                f[i]=f[i-1]+1;
            else f[i]=1;
        }
        initrmq(n,f);
        for(i=1;i<=q;i++)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int t=l;
            while(t<=r&&b[t]==b[t-1])
                t++;
            int cnt=rmq(t,r);
            int ans=max(t-l,cnt);
            printf("%d\n",ans);
        }
    }
    return 0;
}