#include <bits/stdc++.h>
using namespace std;
int l, i, j, n, k, k1, k2, k3, k4, k5, m, fldgjdflgjhrthrl, fldggfhfghjdflgjl,
    fldgjdflgrtyrtyjl, ffgfldgjdflgjl, r, x, y, z, ans;
int a[558228], d1[405], d2[405], p[558228], nxt[558228], prv[558228],
    val[558228];
set<int> f;
set<int>::iterator itr;
pair<int, int> g[558228];
vector<pair<int, int> > gg;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int Abs(int x) { return x > 0 ? x : -x; }
int dsu_get(int v) { return (v == p[v]) ? v : (p[v] = dsu_get(p[v])); }
void dsu_unite(int a, int b) {
  a = dsu_get(a);
  b = dsu_get(b);
  if (rand() & 1) swap(a, b);
  if (a != b) p[a] = b;
}
int main() {
  srand(time(0));
  int q;
  cin >> n >> q;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 1; i < n; i++) {
    gg.push_back(make_pair(Abs(a[i + 1] - a[i]), i));
  }
  sort(gg.begin(), gg.end());
  for (i = 0; i < q; i++) {
    cin >> x >> y;
    int sz = 0;
    for (j = 0; j < n - 1; j++)
      if (gg[j].second >= x && gg[j].second < y) g[sz++] = gg[j];
    long long ans = 0;
    for (j = x; j < y; j++) {
      nxt[j] = j + 1;
      val[j] = 1;
      p[j] = j;
    }
    val[y] = 1;
    p[y] = y;
    for (j = 0; j < sz; j++) {
      int num = dsu_get(g[j].second);
      int val1 = val[num];
      int tmp = dsu_get(nxt[num]);
      int val2 = val[tmp];
      int tmp_val = val[num] + val[tmp];
      nxt[num] = nxt[tmp];
      dsu_unite(num, tmp);
      val[dsu_get(num)] = tmp_val;
      ans += (long long)val1 * val2 * g[j].first;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
int l, i, j, n, k, k1, k2, k3, k4, k5, m, fldgjdflgjhrthrl, fldggfhfghjdflgjl,
    fldgjdflgrtyrtyjl, ffgfldgjdflgjl, r, x, y, z, ans;
int a[558228], d1[405], d2[405], p[558228], nxt[558228], prv[558228],
    val[558228];
set<int> f;
set<int>::iterator itr;
pair<int, int> g[558228];
vector<pair<int, int> > gg;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int Abs(int x) { return x > 0 ? x : -x; }
int dsu_get(int v) { return (v == p[v]) ? v : (p[v] = dsu_get(p[v])); }
void dsu_unite(int a, int b) {
  a = dsu_get(a);
  b = dsu_get(b);
  if (rand() & 1) swap(a, b);
  if (a != b) p[a] = b;
}
int main() {
  srand(time(0));
  int q;
  cin >> n >> q;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 1; i < n; i++) {
    gg.push_back(make_pair(Abs(a[i + 1] - a[i]), i));
  }
  sort(gg.begin(), gg.end());
  for (i = 0; i < q; i++) {
    cin >> x >> y;
    int sz = 0;
    for (j = 0; j < n - 1; j++)
      if (gg[j].second >= x && gg[j].second < y) g[sz++] = gg[j];
    long long ans = 0;
    for (j = x; j < y; j++) {
      nxt[j] = j + 1;
      val[j] = 1;
      p[j] = j;
    }
    val[y] = 1;
    p[y] = y;
    for (j = 0; j < sz; j++) {
      int num = dsu_get(g[j].second);
      int val1 = val[num];
      int tmp = dsu_get(nxt[num]);
      int val2 = val[tmp];
      int tmp_val = val[num] + val[tmp];
      nxt[num] = nxt[tmp];
      dsu_unite(num, tmp);
      val[dsu_get(num)] = tmp_val;
      ans += (long long)val1 * val2 * g[j].first;
    }
    cout << ans << endl;
  }
  return 0;
}

```