#include <bits/stdc++.h>
using namespace std;
const long long sz = 100005, mod = 1000000007;
long long noq, q, n, m, i, j, k, cnt = 0, sum = 0, msum = 0, temp;
long long a[sz];
string s, t;
void solve() {
  vector<pair<long long, long long> > v;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  i = 0;
  while (i <= n) {
    long long si, sj;
    i++;
    si = i;
    while (i <= n and a[i + 1] >= a[i]) i++;
    i++;
    while (i <= n and a[i + 1] <= a[i]) i++;
    sj = i;
    v.push_back({si, sj});
    if (i > n) break;
    while (a[i - 1] <= a[i]) i--;
    i--;
  }
  while (m-- > 0) {
    long long l, r;
    cin >> l >> r;
    long long pos =
        lower_bound(v.begin(), v.end(), make_pair(l, LONG_LONG_MIN)) -
        v.begin();
    if (l == r || r == l + 1)
      cout << "Yes"
           << "\n";
    else if (pos == v.size()) {
      pos--;
      if (v[pos].first <= l and r <= v[pos].second)
        cout << "Yes"
             << "\n";
      else
        cout << "No"
             << "\n";
    } else if (v[pos].first == l) {
      if (r <= v[pos].second)
        cout << "Yes"
             << "\n";
      else
        cout << "No"
             << "\n";
    } else if (pos == 0)
      cout << "No"
           << "\n";
    else {
      pos--;
      if (v[pos].first <= l and r <= v[pos].second)
        cout << "Yes"
             << "\n";
      else
        cout << "No"
             << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  noq = 1;
  while (noq-- > 0) solve();
  return 0;
}

### Prompt

Generate a cpp solution to 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;
const long long sz = 100005, mod = 1000000007;
long long noq, q, n, m, i, j, k, cnt = 0, sum = 0, msum = 0, temp;
long long a[sz];
string s, t;
void solve() {
  vector<pair<long long, long long> > v;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  i = 0;
  while (i <= n) {
    long long si, sj;
    i++;
    si = i;
    while (i <= n and a[i + 1] >= a[i]) i++;
    i++;
    while (i <= n and a[i + 1] <= a[i]) i++;
    sj = i;
    v.push_back({si, sj});
    if (i > n) break;
    while (a[i - 1] <= a[i]) i--;
    i--;
  }
  while (m-- > 0) {
    long long l, r;
    cin >> l >> r;
    long long pos =
        lower_bound(v.begin(), v.end(), make_pair(l, LONG_LONG_MIN)) -
        v.begin();
    if (l == r || r == l + 1)
      cout << "Yes"
           << "\n";
    else if (pos == v.size()) {
      pos--;
      if (v[pos].first <= l and r <= v[pos].second)
        cout << "Yes"
             << "\n";
      else
        cout << "No"
             << "\n";
    } else if (v[pos].first == l) {
      if (r <= v[pos].second)
        cout << "Yes"
             << "\n";
      else
        cout << "No"
             << "\n";
    } else if (pos == 0)
      cout << "No"
           << "\n";
    else {
      pos--;
      if (v[pos].first <= l and r <= v[pos].second)
        cout << "Yes"
             << "\n";
      else
        cout << "No"
             << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  noq = 1;
  while (noq-- > 0) solve();
  return 0;
}

```