Ladder (单调栈)

Ladder

Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

Description

You've got an array, consisting of n integersa₁, a₂, ..., a_n. Also, you've gotm queries, the i-th query is described by two integersl_i, r_i. Numbersl_i, r_i define a subsegment of the original array, that is, the sequence of numbersa_li, a_li + 1, a_li + 2, ..., a_ri. For each query you should check whether the corresponding segment is a ladder.

A ladder is a sequence of integers b₁, b₂, ..., b_k, such that it first doesn't decrease, then doesn't increase. In other words, there is such integerx(1 ≤ x ≤ k), that the following inequation fulfills:b₁ ≤ b₂ ≤ ... ≤ b_x ≥ b_x + 1 ≥ b_x + 2... ≥ b_k. Note that the non-decreasing and the non-increasing sequences are also considered ladders.

Input

The first line contains two integers n andm(1 ≤ n, m ≤ 10⁵) — the number of array elements and the number of queries. The second line contains the sequence of integersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 10⁹), where number a_i stands for thei-th array element.

The following m lines contain the description of the queries. Thei-th line contains the description of thei-th query, consisting of two integers l_i,r_i(1 ≤ l_i ≤ r_i ≤ n) — the boundaries of the subsegment of the initial array.

The numbers in the lines are separated by single spaces.

Output

Print m lines, in the i-th line print word "Yes" (without the quotes), if the subsegment that corresponds to thei-th query is the ladder, or word "No" (without the quotes) otherwise.

Sample Input

Input

8 61 2 1 3 3 5 2 11 32 32 48 81 45 8

Output

YesYesNoYesNoYes题意：给出一个数列，m次询问，每次询问一个区间，判断这个区间是否是按照非递减和非递增变化的分析：对于每一个区间的端点，如果我们知道了左端点和右端点左右各能达到的最大高度的下标，那么我们就能判断这个区间是否是合法的，那么单调栈预处理一下就可以了
#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>//#pragma comment(linker, "/STACK:1024000000,1024000000");using namespace std;#define INF 0x3f3f3f3fint a[100005];int l[100005],r[100005];int s[100005];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            l[i]=r[i]=i;        }        a[0]=-INF;        a[n+1]=-INF;        int p=-1;        for(int i=0;i<=n+1;i++)        {            if(p>=0&&a[s[p]]>a[i])            {                for(int j=0;j<p;j++)                {                    r[s[j]]=s[p];                }                p=-1;            }            s[++p]=i;        }        p=-1;        for(int i=n+1;i>=0;i--)        {            if(p>=0&&a[s[p]]>a[i])            {                for(int j=0;j<p;j++)                {                    l[s[j]]=s[p];                }                p=-1;            }            s[++p]=i;        }        while(m--)        {            int a,b;            scanf("%d%d",&a,&b);            if(l[a]==l[b]&&l[a]<=a)            {                printf("Yes\n");            }            else if(r[a]==r[b]&&r[b]>=b)            {                printf("Yes\n");            }            else if(r[a]==l[b]&&r[a]>=a&&r[a]<=b)            {                printf("Yes\n");            }            else if(r[a]>=l[b])            {                printf("Yes\n");            }            else            {                printf("No\n");            }        }    }    return 0;}