#include <bits/stdc++.h>
using namespace std;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N, Q;
  cin >> N >> Q;
  int prev;
  cin >> prev;
  vector<int> h(N - 1);
  for (int i = 0; i < N - 1; i++) {
    int x;
    cin >> x;
    h[i] = abs(x - prev);
    prev = x;
  }
  N--;
  vector<pair<int, int> > queries(Q);
  for (int i = 0; i < Q; i++) {
    cin >> queries[i].first >> queries[i].second;
    queries[i].first--;
    queries[i].second--;
  }
  vector<int> after(N);
  stack<pair<int, int> > st;
  st.emplace(N, INT_MAX);
  for (int i = N - 1; i >= 0; i--) {
    while (h[i] > st.top().second) st.pop();
    after[i] = st.top().first;
    st.emplace(i, h[i]);
  }
  vector<long long> ans(Q);
  st = stack<pair<int, int> >();
  st.emplace(-1, INT_MAX);
  for (int i = 0; i < N; i++) {
    while (h[i] >= st.top().second) st.pop();
    int before = st.top().first;
    for (int j = 0; j < Q; j++) {
      long long L = i - max(before, queries[j].first - 1);
      long long R = min(after[i], queries[j].second) - i;
      if (L > 0 && R > 0) ans[j] += L * R * h[i];
    }
    st.emplace(i, h[i]);
  }
  for (int i = 0; i < Q; i++) cout << ans[i] << '\n';
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

A function <image> is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all <image>. We'll deal with a more... discrete version of this term.

For an array <image>, we define it's Lipschitz constant <image> as follows:

  * if n < 2, <image>
  * if n ≥ 2, <image> over all 1 ≤ i < j ≤ n



In other words, <image> is the smallest non-negative integer such that |h[i] - h[j]| ≤ L·|i - j| holds for all 1 ≤ i, j ≤ n.

You are given an array <image> of size n and q queries of the form [l, r]. For each query, consider the subarray <image>; determine the sum of Lipschitz constants of all subarrays of <image>.

Input

The first line of the input contains two space-separated integers n and q (2 ≤ n ≤ 100 000 and 1 ≤ q ≤ 100) — the number of elements in array <image> and the number of queries respectively.

The second line contains n space-separated integers <image> (<image>).

The following q lines describe queries. The i-th of those lines contains two space-separated integers li and ri (1 ≤ li < ri ≤ n).

Output

Print the answers to all queries in the order in which they are given in the input. For the i-th query, print one line containing a single integer — the sum of Lipschitz constants of all subarrays of <image>.

Examples

Input

10 4
1 5 2 9 1 3 4 2 1 7
2 4
3 8
7 10
1 9


Output

17
82
23
210


Input

7 6
5 7 7 4 6 6 2
1 2
2 3
2 6
1 7
4 7
3 5


Output

2
0
22
59
16
8

Note

In the first query of the first sample, the Lipschitz constants of subarrays of <image> with length at least 2 are:

  * <image>
  * <image>
  * <image>



The answer to the query is their sum.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N, Q;
  cin >> N >> Q;
  int prev;
  cin >> prev;
  vector<int> h(N - 1);
  for (int i = 0; i < N - 1; i++) {
    int x;
    cin >> x;
    h[i] = abs(x - prev);
    prev = x;
  }
  N--;
  vector<pair<int, int> > queries(Q);
  for (int i = 0; i < Q; i++) {
    cin >> queries[i].first >> queries[i].second;
    queries[i].first--;
    queries[i].second--;
  }
  vector<int> after(N);
  stack<pair<int, int> > st;
  st.emplace(N, INT_MAX);
  for (int i = N - 1; i >= 0; i--) {
    while (h[i] > st.top().second) st.pop();
    after[i] = st.top().first;
    st.emplace(i, h[i]);
  }
  vector<long long> ans(Q);
  st = stack<pair<int, int> >();
  st.emplace(-1, INT_MAX);
  for (int i = 0; i < N; i++) {
    while (h[i] >= st.top().second) st.pop();
    int before = st.top().first;
    for (int j = 0; j < Q; j++) {
      long long L = i - max(before, queries[j].first - 1);
      long long R = min(after[i], queries[j].second) - i;
      if (L > 0 && R > 0) ans[j] += L * R * h[i];
    }
    st.emplace(i, h[i]);
  }
  for (int i = 0; i < Q; i++) cout << ans[i] << '\n';
  return 0;
}

```