#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000009;
const int MAXN = 100007;
int treeMax[4 * MAXN], treeMin[4 * MAXN];
void UpdateMax(int a, int val) {
  a += MAXN;
  treeMax[a] = val;
  while (a) {
    treeMax[a] = max(treeMax[a], val);
    a /= 2;
  }
}
void UpdateMin(int a, int val) {
  a += MAXN;
  treeMin[a] = val;
  while (a) {
    treeMin[a] = min(treeMin[a], val);
    a /= 2;
  }
}
int GetMax(int a, int b) {
  a += MAXN;
  b += MAXN;
  int ret = 0;
  while (a <= b) {
    if (a % 2 == 1) ret = max(ret, treeMax[a]);
    if (b % 2 == 0) ret = max(ret, treeMax[b]);
    a = (a + 1) / 2;
    b = (b - 1) / 2;
  }
  return ret;
}
int GetMin(int a, int b) {
  a += MAXN;
  b += MAXN;
  int ret = INF;
  while (a <= b) {
    if (a % 2 == 1) ret = min(ret, treeMin[a]);
    if (b % 2 == 0) ret = min(ret, treeMin[b]);
    a = (a + 1) / 2;
    b = (b - 1) / 2;
  }
  return ret;
}
int n;
int a[MAXN + 3], tab[MAXN + 3];
pair<int, int> prep[MAXN + 3];
int Left[MAXN + 3], Right[MAXN + 3];
void Preprocessing() {
  for (int i = 1; i <= n; i++) prep[i] = make_pair(tab[i], i);
  sort(prep + 1, prep + n + 1);
  for (int i = n; i >= 1; i--) {
    Left[prep[i].second] = GetMax(1, prep[i].second);
    Right[prep[i].second] = GetMin(prep[i].second, n);
    UpdateMax(prep[i].second, prep[i].second);
    UpdateMin(prep[i].second, prep[i].second);
  }
}
long long Solve(int l, int r) {
  long long ret = 0;
  for (int i = l; i <= r; i++) {
    int first = i - max(l - 1, Left[i]);
    int second = min(r + 1, Right[i]) - i;
    ret += (long long)first * second * tab[i];
  }
  return ret;
}
int main() {
  int t;
  scanf("%d %d", &n, &t);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) tab[i] = abs(a[i] - a[i + 1]);
  n--;
  for (int i = 0; i < 4 * MAXN; i++) treeMin[i] = n + 1;
  Preprocessing();
  while (t--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%lld\n", Solve(l, r - 1));
  }
  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 int INF = 1000000009;
const int MAXN = 100007;
int treeMax[4 * MAXN], treeMin[4 * MAXN];
void UpdateMax(int a, int val) {
  a += MAXN;
  treeMax[a] = val;
  while (a) {
    treeMax[a] = max(treeMax[a], val);
    a /= 2;
  }
}
void UpdateMin(int a, int val) {
  a += MAXN;
  treeMin[a] = val;
  while (a) {
    treeMin[a] = min(treeMin[a], val);
    a /= 2;
  }
}
int GetMax(int a, int b) {
  a += MAXN;
  b += MAXN;
  int ret = 0;
  while (a <= b) {
    if (a % 2 == 1) ret = max(ret, treeMax[a]);
    if (b % 2 == 0) ret = max(ret, treeMax[b]);
    a = (a + 1) / 2;
    b = (b - 1) / 2;
  }
  return ret;
}
int GetMin(int a, int b) {
  a += MAXN;
  b += MAXN;
  int ret = INF;
  while (a <= b) {
    if (a % 2 == 1) ret = min(ret, treeMin[a]);
    if (b % 2 == 0) ret = min(ret, treeMin[b]);
    a = (a + 1) / 2;
    b = (b - 1) / 2;
  }
  return ret;
}
int n;
int a[MAXN + 3], tab[MAXN + 3];
pair<int, int> prep[MAXN + 3];
int Left[MAXN + 3], Right[MAXN + 3];
void Preprocessing() {
  for (int i = 1; i <= n; i++) prep[i] = make_pair(tab[i], i);
  sort(prep + 1, prep + n + 1);
  for (int i = n; i >= 1; i--) {
    Left[prep[i].second] = GetMax(1, prep[i].second);
    Right[prep[i].second] = GetMin(prep[i].second, n);
    UpdateMax(prep[i].second, prep[i].second);
    UpdateMin(prep[i].second, prep[i].second);
  }
}
long long Solve(int l, int r) {
  long long ret = 0;
  for (int i = l; i <= r; i++) {
    int first = i - max(l - 1, Left[i]);
    int second = min(r + 1, Right[i]) - i;
    ret += (long long)first * second * tab[i];
  }
  return ret;
}
int main() {
  int t;
  scanf("%d %d", &n, &t);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) tab[i] = abs(a[i] - a[i + 1]);
  n--;
  for (int i = 0; i < 4 * MAXN; i++) treeMin[i] = n + 1;
  Preprocessing();
  while (t--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%lld\n", Solve(l, r - 1));
  }
  return 0;
}

```