CQUOJ 9920 Ladder

You've got an array, consisting of n integers a₁, a₂, ..., a_n. Also, you've got m queries, the i-th query is described by two integers l_i, r_i. Numbers l_i, r_i define a subsegment of the original array, that is, the sequence of numbers a_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 integer x (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 and m (1 ≤ n, m ≤ 10⁵) — the number of array elements and the number of queries. The second line contains the sequence of integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where number a_i stands for the i-th array element.

The following m lines contain the description of the queries. The i-th line contains the description of the i-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 the i-th query is the ladder, or word "No" (without the quotes) otherwise.

 

Sample Input

Input

8 6
1 2 1 3 3 5 2 1
1 3
2 3
2 4
8 8
1 4
5 8

Output

Yes
Yes
No
Yes
No
Yes

/*
2016年4月22日21:37:37 
    题意: 给出一个序列，m次查询，每次给出一个子串，
        问这个子串是否满足，中间能够找到一个元素，
        让这个元素作为前后分别单调的分界. 
    预处理 向左向右分别扫一遍,记录某一个点向右或向左 离的最近的的那个最大数的位置 
    然后就可以判断了 好好想想 
    样例  1 2 1 4 5 5 2 1
    样例  1 2 1 4 5 4 5 1
*/
# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
# include <vector>
# include <cmath>
# define INF 0x3f3f3f3f
using namespace std;
const int N = 1e5;
int a[N], R[N], L[N];

int main(void)
{
    int n, m, i, l, r;
    while (~scanf("%d %d", &n, &m))
    {
        for (i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        R[n] = n;
        for (i = n-1; i >= 1; i--){
            if (a[i] > a[i+1]) R[i] = i;
            else R[i] = R[i+1];
        }
        L[1] = 1;
        for (i = 2; i <= n; i++){
            if (a[i] > a[i-1]) L[i] = i;
            else L[i] = L[i-1];
        }
        while (m--){
            scanf("%d %d", &l, &r);
            if (L[r] <= R[l])        
                printf("Yes\n");     
            else printf("No\n");     
        }
    }
    
    return 0;    
}