#include <bits/stdc++.h>
using namespace std;
int len;
long long int a[400005], d[400005];
long long int m[400005];
int better(int i, int j) { return d[i] > d[j] ? i : j; }
void pre() {
  int i;
  for (i = 0; i < len; ++i) m[len + i] = i;
  for (i = len - 1; i >= 1; --i) m[i] = better(m[i << 1], m[i << 1 | 1]);
}
int best(int l, int r) {
  int kl = l, kr = r - 1;
  for (l += len, r += len; l < r; l >>= 1, r >>= 1) {
    if (l & 1) kl = better(kl, m[l++]);
    if (r & 1) kr = better(kr, m[--r]);
  }
  return better(kl, kr);
}
long long int g(int lo, int hi) {
  long long int ans;
  int k;
  if (hi - lo == 0)
    ans = 0;
  else if (hi - lo == 1)
    ans = d[lo];
  else
    k = best(lo, hi),
    ans = (hi - k) * d[k] * (k - lo + 1) + g(lo, k) + g(k + 1, hi);
  return ans;
}
int main() {
  long long int ans;
  int i, l, r, Q;
  while (scanf("%d%d", &len, &Q) == 2) {
    for (i = 0; i < len; ++i) scanf("%lld", &a[i]);
    int newlen = 1;
    while (newlen < len) newlen <<= 1;
    for (i = len; i < newlen; ++i) a[i] = 0;
    len = newlen;
    for (i = 0; i < len; ++i) d[i] = abs(a[i + 1] - a[i]);
    pre();
    while (Q--) {
      scanf("%d%d", &l, &r), --l, --r;
      ans = g(l, r);
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Please formulate 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 len;
long long int a[400005], d[400005];
long long int m[400005];
int better(int i, int j) { return d[i] > d[j] ? i : j; }
void pre() {
  int i;
  for (i = 0; i < len; ++i) m[len + i] = i;
  for (i = len - 1; i >= 1; --i) m[i] = better(m[i << 1], m[i << 1 | 1]);
}
int best(int l, int r) {
  int kl = l, kr = r - 1;
  for (l += len, r += len; l < r; l >>= 1, r >>= 1) {
    if (l & 1) kl = better(kl, m[l++]);
    if (r & 1) kr = better(kr, m[--r]);
  }
  return better(kl, kr);
}
long long int g(int lo, int hi) {
  long long int ans;
  int k;
  if (hi - lo == 0)
    ans = 0;
  else if (hi - lo == 1)
    ans = d[lo];
  else
    k = best(lo, hi),
    ans = (hi - k) * d[k] * (k - lo + 1) + g(lo, k) + g(k + 1, hi);
  return ans;
}
int main() {
  long long int ans;
  int i, l, r, Q;
  while (scanf("%d%d", &len, &Q) == 2) {
    for (i = 0; i < len; ++i) scanf("%lld", &a[i]);
    int newlen = 1;
    while (newlen < len) newlen <<= 1;
    for (i = len; i < newlen; ++i) a[i] = 0;
    len = newlen;
    for (i = 0; i < len; ++i) d[i] = abs(a[i + 1] - a[i]);
    pre();
    while (Q--) {
      scanf("%d%d", &l, &r), --l, --r;
      ans = g(l, r);
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```