#include <bits/stdc++.h>
using namespace std;
template <class Tp>
Tp mymin(Tp x, Tp y) {
  return x < y ? x : y;
}
template <class Tp>
Tp mymax(Tp x, Tp y) {
  return x < y ? y : x;
}
template <class Tp>
Tp myabs(Tp x) {
  return x < 0 ? -x : x;
}
template <class Tp>
void myswap(Tp &x, Tp &y) {
  Tp z = x;
  x = y, y = z;
}
template <class Tp>
Tp read() {
  Tp res = 0, f_f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f_f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) - '0' + ch, ch = getchar();
  return res * f_f;
}
const int maxn = 100005;
int n, q, a[maxn], b[maxn], L[maxn], R[maxn];
int stac[maxn], topc;
int main() {
  n = (read<int>()), q = (read<int>());
  for (int i = 1; i <= n; i++) {
    a[i] = (read<int>());
    if (i ^ 1) b[i - 1] = myabs(a[i] - a[i - 1]);
  }
  for (int i = 1; i <= n - 1; i++) {
    while (topc && b[stac[topc]] <= b[i]) topc--;
    if (!topc)
      L[i] = 1;
    else
      L[i] = stac[topc] + 1;
    stac[++topc] = i;
  }
  topc = 0;
  for (int i = n - 1; i >= 1; i--) {
    while (topc && b[stac[topc]] < b[i]) topc--;
    if (!topc)
      R[i] = n - 1;
    else
      R[i] = stac[topc] - 1;
    stac[++topc] = i;
  }
  while (q--) {
    int l = (read<int>()), r = (read<int>()) - 1;
    long long res = 0;
    for (int i = l; i <= r; i++) {
      int x = L[i], y = R[i];
      x = mymax(x, l), y = mymin(y, r);
      res += 1ll * (y - i + 1) * (i - x + 1) * b[i];
    }
    printf("%lld\n", res);
  }
  return 0;
}

### Prompt

Generate 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;
template <class Tp>
Tp mymin(Tp x, Tp y) {
  return x < y ? x : y;
}
template <class Tp>
Tp mymax(Tp x, Tp y) {
  return x < y ? y : x;
}
template <class Tp>
Tp myabs(Tp x) {
  return x < 0 ? -x : x;
}
template <class Tp>
void myswap(Tp &x, Tp &y) {
  Tp z = x;
  x = y, y = z;
}
template <class Tp>
Tp read() {
  Tp res = 0, f_f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f_f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) - '0' + ch, ch = getchar();
  return res * f_f;
}
const int maxn = 100005;
int n, q, a[maxn], b[maxn], L[maxn], R[maxn];
int stac[maxn], topc;
int main() {
  n = (read<int>()), q = (read<int>());
  for (int i = 1; i <= n; i++) {
    a[i] = (read<int>());
    if (i ^ 1) b[i - 1] = myabs(a[i] - a[i - 1]);
  }
  for (int i = 1; i <= n - 1; i++) {
    while (topc && b[stac[topc]] <= b[i]) topc--;
    if (!topc)
      L[i] = 1;
    else
      L[i] = stac[topc] + 1;
    stac[++topc] = i;
  }
  topc = 0;
  for (int i = n - 1; i >= 1; i--) {
    while (topc && b[stac[topc]] < b[i]) topc--;
    if (!topc)
      R[i] = n - 1;
    else
      R[i] = stac[topc] - 1;
    stac[++topc] = i;
  }
  while (q--) {
    int l = (read<int>()), r = (read<int>()) - 1;
    long long res = 0;
    for (int i = l; i <= r; i++) {
      int x = L[i], y = R[i];
      x = mymax(x, l), y = mymin(y, r);
      res += 1ll * (y - i + 1) * (i - x + 1) * b[i];
    }
    printf("%lld\n", res);
  }
  return 0;
}

```