#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, q, a[N], b[N];
int st[N], top = 0;
long long ans[N];
struct BIT {
  long long s1[N], s[N];
  void add(long long s[], int first, long long d) {
    for (; first <= n + 1; first += (-first) & first) s[first] += d;
  }
  long long sum(long long s[], int first) {
    long long ans = 0;
    for (; first; first -= (-first) & first) ans += s[first];
    return ans;
  }
  void Add(int first, long long d) {
    add(s, first, d), add(s1, first, 1LL * d * first);
  }
  long long Sum(int first) {
    return 1LL * sum(s, first) * (first + 1) - 1LL * sum(s1, first);
  }
  void Add(int l, int r, long long d) { Add(l, d), Add(r + 1, -d); }
  long long Sum(int l, int r) { return Sum(r) - Sum(l - 1); }
} s, s1;
vector<pair<int, int> > Q[N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    r--;
    Q[r].push_back(make_pair(l, i));
  }
  n--;
  for (int i = 1; i <= n; i++)
    b[i] = ((a[i] - a[i + 1]) < 0 ? -(a[i] - a[i + 1]) : (a[i] - a[i + 1]));
  for (int i = 1; i <= n; i++) {
    while (top && b[i] >= b[st[top]]) {
      s.Add(st[top - 1] + 1, st[top], 1LL * b[st[top]] * (i - 1));
      s1.Add(st[top - 1] + 1, st[top], -b[st[top]]);
      top--;
    }
    s.Add(st[top] + 1, i, -1LL * b[i] * (i - 1));
    s1.Add(st[top] + 1, i, b[i]);
    st[++top] = i;
    int len = Q[i].size();
    for (int j = 0; j < len; j++)
      ans[Q[i][j].second] =
          s.Sum(Q[i][j].first, i) + s1.Sum(Q[i][j].first, i) * i;
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  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 int N = 1e5 + 10;
int n, q, a[N], b[N];
int st[N], top = 0;
long long ans[N];
struct BIT {
  long long s1[N], s[N];
  void add(long long s[], int first, long long d) {
    for (; first <= n + 1; first += (-first) & first) s[first] += d;
  }
  long long sum(long long s[], int first) {
    long long ans = 0;
    for (; first; first -= (-first) & first) ans += s[first];
    return ans;
  }
  void Add(int first, long long d) {
    add(s, first, d), add(s1, first, 1LL * d * first);
  }
  long long Sum(int first) {
    return 1LL * sum(s, first) * (first + 1) - 1LL * sum(s1, first);
  }
  void Add(int l, int r, long long d) { Add(l, d), Add(r + 1, -d); }
  long long Sum(int l, int r) { return Sum(r) - Sum(l - 1); }
} s, s1;
vector<pair<int, int> > Q[N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    r--;
    Q[r].push_back(make_pair(l, i));
  }
  n--;
  for (int i = 1; i <= n; i++)
    b[i] = ((a[i] - a[i + 1]) < 0 ? -(a[i] - a[i + 1]) : (a[i] - a[i + 1]));
  for (int i = 1; i <= n; i++) {
    while (top && b[i] >= b[st[top]]) {
      s.Add(st[top - 1] + 1, st[top], 1LL * b[st[top]] * (i - 1));
      s1.Add(st[top - 1] + 1, st[top], -b[st[top]]);
      top--;
    }
    s.Add(st[top] + 1, i, -1LL * b[i] * (i - 1));
    s1.Add(st[top] + 1, i, b[i]);
    st[++top] = i;
    int len = Q[i].size();
    for (int j = 0; j < len; j++)
      ans[Q[i][j].second] =
          s.Sum(Q[i][j].first, i) + s1.Sum(Q[i][j].first, i) * i;
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  return 0;
}

```