#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238463;
const long long mod = 1000000007;
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U> &l, const std::pair<T, U> &r) {
  return {l.first + r.first, l.second + r.second};
}
struct E {
  int x;
  int y;
  int z;
};
bool cmpf(E &a1, E &a2) {
  if (a1.x != a2.x) return a1.x < a2.x;
  if (a1.y != a2.y) return a1.y < a2.y;
  return a1.z < a2.z;
}
const int N = 100001;
int n, q, l, r;
int x[N];
int a[N];
long long u[N], v[N];
int main() {
  cin >> n >> q;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n - 1); ++i) x[i] = abs(a[i + 1] - a[i]);
  for (int k = (1); k <= (q); ++k) {
    scanf("%d%d", &l, &r);
    --r;
    stack<int> S, T;
    for (int i = (l); i <= (r); ++i) {
      if (S.size())
        while (S.size() && x[S.top()] < x[i]) S.pop();
      if (S.empty())
        u[i] = i - l + 1;
      else
        u[i] = i - S.top();
      S.push(i);
    }
    for (int i = (r); i >= (l); --i) {
      if (T.size())
        while (T.size() && x[T.top()] <= x[i]) T.pop();
      if (T.empty())
        v[i] = r - i + 1;
      else
        v[i] = T.top() - i;
      T.push(i);
    }
    long long ans = 0;
    for (int i = (l); i <= (r); ++i) ans += (u[i] * v[i]) * x[i];
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 double PI = 3.141592653589793238463;
const long long mod = 1000000007;
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U> &l, const std::pair<T, U> &r) {
  return {l.first + r.first, l.second + r.second};
}
struct E {
  int x;
  int y;
  int z;
};
bool cmpf(E &a1, E &a2) {
  if (a1.x != a2.x) return a1.x < a2.x;
  if (a1.y != a2.y) return a1.y < a2.y;
  return a1.z < a2.z;
}
const int N = 100001;
int n, q, l, r;
int x[N];
int a[N];
long long u[N], v[N];
int main() {
  cin >> n >> q;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n - 1); ++i) x[i] = abs(a[i + 1] - a[i]);
  for (int k = (1); k <= (q); ++k) {
    scanf("%d%d", &l, &r);
    --r;
    stack<int> S, T;
    for (int i = (l); i <= (r); ++i) {
      if (S.size())
        while (S.size() && x[S.top()] < x[i]) S.pop();
      if (S.empty())
        u[i] = i - l + 1;
      else
        u[i] = i - S.top();
      S.push(i);
    }
    for (int i = (r); i >= (l); --i) {
      if (T.size())
        while (T.size() && x[T.top()] <= x[i]) T.pop();
      if (T.empty())
        v[i] = r - i + 1;
      else
        v[i] = T.top() - i;
      T.push(i);
    }
    long long ans = 0;
    for (int i = (l); i <= (r); ++i) ans += (u[i] * v[i]) * x[i];
    printf("%I64d\n", ans);
  }
  return 0;
}

```