#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
struct Input {
  int n, q;
  vector<int> a, l, r;
  bool read() {
    if (scanf("%d%d", &n, &q) != 2) return 0;
    a.resize(n);
    for (int i = int(0); i < int(n); ++i) scanf("%d", &a[i]);
    l.resize(q);
    r = l;
    for (int i = int(0); i < int(q); ++i) {
      scanf("%d%d", &l[i], &r[i]);
      --l[i], --r[i];
    }
    return 1;
  }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  vector<int64_t> ans;
  void write() {
    for (int64_t x : ans) {
      printf("%" PRId64 "\n", x);
    }
  }
};
struct Solution : Data {
  vector<int> h;
  static const int MAXL = 17;
  vector<vector<pair<int, int> > > sp;
  vector<int> loglen;
  int get_max_id(int l, int r) {
    if (l == r) {
      return l;
    }
    int it = loglen[r - l + 1];
    auto res = max(sp[it][l], sp[it][r - (1 << it) + 1]);
    return res.second;
  }
  int64_t get_ans(int l, int r) {
    if (l >= r) {
      return 0ll;
    }
    int pos = get_max_id(l, r - 1);
    int64_t res = 1ll * h[pos] * (pos - l + 1) * (r - pos);
    res += get_ans(l, pos);
    res += get_ans(pos + 1, r);
    return res;
  }
  void solve() {
    h.resize(n - 1);
    for (int i = int(0); i < int(n - 1); ++i) {
      h[i] = abs(a[i + 1] - a[i]);
    }
    loglen.assign(n, 0);
    int high = 0;
    for (int i = int(2); i < int(n); ++i) {
      loglen[i] = high;
      if ((1 << (high + 1)) == i) ++high;
    }
    sp.assign(MAXL, vector<pair<int, int> >(n - 1, pair<int, int>(-1, 0)));
    for (int i = int(0); i < int(n - 1); ++i) {
      sp[0][i] = make_pair(h[i], i);
    }
    for (int it = int(1); it < int(MAXL); ++it) {
      int pos = (1 << (it - 1));
      sp[it] = sp[it - 1];
      for (int i = 0; pos < n - 1; ++i, ++pos) {
        umx(sp[it][i], sp[it - 1][pos]);
      }
    }
    for (int i = int(0); i < int(q); ++i) {
      ans.push_back(get_ans(l[i], r[i]));
    }
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const long double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
struct Input {
  int n, q;
  vector<int> a, l, r;
  bool read() {
    if (scanf("%d%d", &n, &q) != 2) return 0;
    a.resize(n);
    for (int i = int(0); i < int(n); ++i) scanf("%d", &a[i]);
    l.resize(q);
    r = l;
    for (int i = int(0); i < int(q); ++i) {
      scanf("%d%d", &l[i], &r[i]);
      --l[i], --r[i];
    }
    return 1;
  }
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  vector<int64_t> ans;
  void write() {
    for (int64_t x : ans) {
      printf("%" PRId64 "\n", x);
    }
  }
};
struct Solution : Data {
  vector<int> h;
  static const int MAXL = 17;
  vector<vector<pair<int, int> > > sp;
  vector<int> loglen;
  int get_max_id(int l, int r) {
    if (l == r) {
      return l;
    }
    int it = loglen[r - l + 1];
    auto res = max(sp[it][l], sp[it][r - (1 << it) + 1]);
    return res.second;
  }
  int64_t get_ans(int l, int r) {
    if (l >= r) {
      return 0ll;
    }
    int pos = get_max_id(l, r - 1);
    int64_t res = 1ll * h[pos] * (pos - l + 1) * (r - pos);
    res += get_ans(l, pos);
    res += get_ans(pos + 1, r);
    return res;
  }
  void solve() {
    h.resize(n - 1);
    for (int i = int(0); i < int(n - 1); ++i) {
      h[i] = abs(a[i + 1] - a[i]);
    }
    loglen.assign(n, 0);
    int high = 0;
    for (int i = int(2); i < int(n); ++i) {
      loglen[i] = high;
      if ((1 << (high + 1)) == i) ++high;
    }
    sp.assign(MAXL, vector<pair<int, int> >(n - 1, pair<int, int>(-1, 0)));
    for (int i = int(0); i < int(n - 1); ++i) {
      sp[0][i] = make_pair(h[i], i);
    }
    for (int it = int(1); it < int(MAXL); ++it) {
      int pos = (1 << (it - 1));
      sp[it] = sp[it - 1];
      for (int i = 0; pos < n - 1; ++i, ++pos) {
        umx(sp[it][i], sp[it - 1][pos]);
      }
    }
    for (int i = int(0); i < int(q); ++i) {
      ans.push_back(get_ans(l[i], r[i]));
    }
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}

```