codeforces 279C Ladder（数据技巧）

You've got an array, consisting of n integers a1, a2, ..., an.
Also, you've got mqueries, the i-th query is described by two integers li, ri.
Numbers li, ri define a subsegment of the original array, that is, the sequence of numbers ali, ali + 1, ali + 2, ..., ari.
For each query you should check whether the corresponding segment is a ladder.

A ladder is a sequence of integers b1, b2, ..., bk, 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 ≤ ... ≤ bx ≥ bx + 1 ≥ bx + 2... ≥ bk.
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, ..., an (1 ≤ ai ≤ 109),
where number ai 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 li, ri (1 ≤ li ≤ ri ≤ 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.

Example

Input

8 6
1 2 1 3 3 5 2 1
1 3
2 3
2 4
8 8
1 4
5 8
cc44

Output

Yes
Yes
No
Yes
No
Yes

 【题解】

 这个题刚开始直接跳了，后来做的时候本来想用三分法找山峰，但是TLE了，所以又开始想别的方法，观察了好久终于发现可以双向扫，标记出每个数向左向右能延伸的最大长度，遇到比本身大的就停止计数，一直扫完为止，最后输入查询区间后只要判断左端点是不是小于右端点的计数值就好。

 【AC代码】

# 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
, R
, L
;

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;
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;
}