#include <bits/stdc++.h>
using i64 = long long;
const int N = 1e5 + 9;
int n, m, a[N], v[N], t[N], tot, ls[N], L[N], R[N];
struct Fenwick {
  int n, init;
  std::vector<int> s;
  Fenwick(int n, int init = 0) : n(n), init(init), s(n + 1, init) {}
  void update(int x, int k) {
    if (init != 0) {
      while (x <= n) {
        s[x] = s[x] < k ? s[x] : k;
        x += (x) & (-x);
      }
      return;
    }
    while (x <= n) {
      s[x] = s[x] > k ? s[x] : k;
      x += (x) & (-x);
    }
  }
  int query(int x) {
    if (init != 0) {
      int res = init;
      while (x) {
        res = res < s[x] ? res : s[x];
        x -= (x) & (-x);
      }
      return res;
    }
    int res = 0;
    while (x) {
      res = res > s[x] ? res : s[x];
      x -= (x) & (-x);
    }
    return res;
  }
};
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  std::cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    std::cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    t[i] = v[i] = abs(a[i] - a[i + 1]);
  }
  std::sort(t + 1, t + n);
  tot = std::unique(t + 1, t + n) - t - 1;
  for (int i = 1; i < n; i++) {
    ls[i] = std::lower_bound(t + 1, t + 1 + tot, v[i]) - t;
    ls[i] = tot - ls[i] + 1;
  }
  Fenwick fmx(tot, 0), fmn(tot, n);
  for (int i = 1; i < n; i++) {
    L[i] = fmx.query(ls[i]);
    fmx.update(ls[i], i);
  }
  for (int i = n - 1; i >= 1; i--) {
    R[i] = fmn.query(ls[i] - 1);
    fmn.update(ls[i], i);
  }
  while (m--) {
    int l, r;
    std::cin >> l >> r;
    i64 ans = 0;
    for (int i = l; i < r; i++) {
      int p1 = std::max(L[i], l - 1);
      int p2 = std::min(R[i], r);
      ans += (i64)v[i] * (i - p1) * (p2 - i);
    }
    std::cout << ans << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 i64 = long long;
const int N = 1e5 + 9;
int n, m, a[N], v[N], t[N], tot, ls[N], L[N], R[N];
struct Fenwick {
  int n, init;
  std::vector<int> s;
  Fenwick(int n, int init = 0) : n(n), init(init), s(n + 1, init) {}
  void update(int x, int k) {
    if (init != 0) {
      while (x <= n) {
        s[x] = s[x] < k ? s[x] : k;
        x += (x) & (-x);
      }
      return;
    }
    while (x <= n) {
      s[x] = s[x] > k ? s[x] : k;
      x += (x) & (-x);
    }
  }
  int query(int x) {
    if (init != 0) {
      int res = init;
      while (x) {
        res = res < s[x] ? res : s[x];
        x -= (x) & (-x);
      }
      return res;
    }
    int res = 0;
    while (x) {
      res = res > s[x] ? res : s[x];
      x -= (x) & (-x);
    }
    return res;
  }
};
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  std::cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    std::cin >> a[i];
  }
  for (int i = 1; i < n; i++) {
    t[i] = v[i] = abs(a[i] - a[i + 1]);
  }
  std::sort(t + 1, t + n);
  tot = std::unique(t + 1, t + n) - t - 1;
  for (int i = 1; i < n; i++) {
    ls[i] = std::lower_bound(t + 1, t + 1 + tot, v[i]) - t;
    ls[i] = tot - ls[i] + 1;
  }
  Fenwick fmx(tot, 0), fmn(tot, n);
  for (int i = 1; i < n; i++) {
    L[i] = fmx.query(ls[i]);
    fmx.update(ls[i], i);
  }
  for (int i = n - 1; i >= 1; i--) {
    R[i] = fmn.query(ls[i] - 1);
    fmn.update(ls[i], i);
  }
  while (m--) {
    int l, r;
    std::cin >> l >> r;
    i64 ans = 0;
    for (int i = l; i < r; i++) {
      int p1 = std::max(L[i], l - 1);
      int p2 = std::min(R[i], r);
      ans += (i64)v[i] * (i - p1) * (p2 - i);
    }
    std::cout << ans << "\n";
  }
  return 0;
}

```