#include <bits/stdc++.h>
using namespace std;
int x[100100] = {0}, y[100100] = {0}, z[100100] = {0};
int main() {
  int n, m, i, q, f;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
  }
  i = 1;
  q = 0;
  while (i <= n) {
    q++;
    while (i <= n && x[i] >= x[i - 1]) {
      y[i] = q;
      i++;
    }
    while (i <= n && x[i] <= x[i - 1]) {
      y[i] = q;
      i++;
    }
    f = i - 2;
    z[f + 1] = 1;
    while (f > 0 && x[f] == x[f + 1]) {
      z[f] = 1;
      f--;
    }
  }
  for (i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (y[l] == y[r] || (y[l] + 1 == y[r] && z[l] == 1)) {
      printf("Yes\n");
    } else {
      printf("No\n");
    }
  }
}

### Prompt

Create a solution in CPP for the following problem:

You've got an array, consisting of n integers a1, a2, ..., an. Also, you've got m queries, 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. 

Examples

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int x[100100] = {0}, y[100100] = {0}, z[100100] = {0};
int main() {
  int n, m, i, q, f;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
  }
  i = 1;
  q = 0;
  while (i <= n) {
    q++;
    while (i <= n && x[i] >= x[i - 1]) {
      y[i] = q;
      i++;
    }
    while (i <= n && x[i] <= x[i - 1]) {
      y[i] = q;
      i++;
    }
    f = i - 2;
    z[f + 1] = 1;
    while (f > 0 && x[f] == x[f + 1]) {
      z[f] = 1;
      f--;
    }
  }
  for (i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (y[l] == y[r] || (y[l] + 1 == y[r] && z[l] == 1)) {
      printf("Yes\n");
    } else {
      printf("No\n");
    }
  }
}

```