/*
RMQ问题
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用数组:
cnt[i]:第i段中数的个数（每一段是指所有数都同的一段数）（在这里该数组相当于RMQ问题中的A[]数组）
num[i]:位置i所在段的编号
left1[i]:位置i左端点的位置
right1[i]:位置i右端点的位置
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right1[L]-L+1:从L到该段末尾元素的个数
R-left[R]+1:从该段开始元素到R处的元素个数
RMQ(num[L]+1, num[R]-1):从num[L]+1段到num[R]-1段中的最大cnt;
最后的结果为以上三个数的最大值：
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若：num[L] == num[R]（即：L和R在同一段中）,结果为R-L+1；
----------------------------------------------------------------------------------------------
RMQ：
void RMQ_init()预处理
{
    int n = size;
    for(int i=0; i<n; i++)
        d[i][0] = cnt[i];
    for(int j=1; (1<<j) <= n; j++)
    {
        for(int i=0; i + (1<<j) - 1 < n; i++)
        {
            d[i][j] = max(d[i][j-1], d[i + (1<<(j-1))][j-1]);
        }
    }
}

int RMQ(int L, int R)查询
{
    if(L > R)
        return 0;
    int k = 0;
    while((1<<(k+1)) <= R-L+1)
        k++;
    return max(d[L][k], d[R-(1<<k)+1][k]);
}
--------------------------------------------------------------------------------------
cnt数组，left1数组，right1数组的构建：
int x,p = 100000000;----------------------------P要大于x的取值范围
int ct = -1;------------------------------------输入数据的段数
int _left;--------------------------------------当前段的左端点
for(int i=0; i<n; i++)
{
    scanf("%d",&x);
    if(x == p)
    {
        cnt[ct]++;------------------------------如果输入的数和前一个数同，该段个数+1
        num[i] = ct;----------------------------不断更新位置i所在的段
        left1[i] = _left;-----------------------更左端点
    }
    else
    {
        if(ct >= 0)
        {
            for(int j=_left; j<=i-1; j++)-------更新右端点
                right1[j] = i-1;
        }
        ct++;
        _left = i;
        num[i] = ct;
        left1[i] = _left;
        cnt[ct] = 1;
        p = x;
    }
}
for(int i=_left; i<n; i++)
    right1[i] = n-1;
size = ct+1;
------------------------------------------------------------------------------
*/
#include <iostream>
#include <cstdio>
#include <climits>
#include <cstring>

using namespace std;

const int N = 100010;

int n,q;
int cnt[N],num[N],left1[N],right1[N];
int a[N];
int size;
int d[N][30];

void RMQ_init()
{
    int n = size;
    for(int i=0; i<n; i++)
        d[i][0] = cnt[i];
    for(int j=1; (1<<j) <= n; j++)
    {
        for(int i=0; i + (1<<j) - 1 < n; i++)
        {
            d[i][j] = max(d[i][j-1], d[i + (1<<(j-1))][j-1]);
        }
    }
}

int RMQ(int L, int R)
{
    if(L > R)
        return 0;
    int k = 0;
    while((1<<(k+1)) <= R-L+1)
        k++;
    return max(d[L][k], d[R-(1<<k)+1][k]);
}

int main()
{
    //freopen("input.txt","r",stdin);
    while(scanf("%d%d",&n,&q) != EOF && n)
    {
        memset(cnt, 0, sizeof(cnt));
        memset(left1, 0, sizeof(left1));
        memset(right1, 0, sizeof(right1));
        memset(d, 0, sizeof(d));
        int x,p = 100000000;
        int ct = -1;
        int _left;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&x);
            if(x == p)
            {
                cnt[ct]++;
                num[i] = ct;
                left1[i] = _left;
            }
            else
            {
                if(ct >= 0)
                {
                    for(int j=_left; j<=i-1; j++)
                        right1[j] = i-1;
                }
                ct++;
                _left = i;
                num[i] = ct;
                left1[i] = _left;
                cnt[ct] = 1;
                p = x;
            }
        }
        for(int i=_left; i<n; i++)
            right1[i] = n-1;
        size = ct+1;
        RMQ_init();
        while(q--)
        {
            int L,R;
            scanf("%d%d",&L,&R);
            L--;
            R--;
            int t1 = right1[L]-L+1;
            int t2 = RMQ(num[L]+1, num[R]-1);
            int t3 = R-left1[R]+1;
            if(num[L] == num[R])
                printf("%d\n",R-L+1);
            else
                printf("%d\n",max(t1, max(t2, t3)));
        }
    }
    return 0;
}