#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
struct Node {
  int pos, val;
  Node() {}
  Node(int pos, int val) : pos(pos), val(val) {}
} st[maxn];
struct Query {
  int l, r;
  bool operator<(const Query& tmp) const { return r < tmp.r; }
} query[maxn];
int n, q, top;
int a[maxn], l[maxn], r[maxn];
long long ans[maxn];
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
int main() {
  if (fopen("in.in", "r") != NULL) {
    freopen("in.in", "r", stdin);
    freopen("out.out", "w", stdout);
  }
  n = read(), q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = n; i >= 1; i--) a[i] = abs(a[i] - a[i - 1]);
  for (int i = 2; i <= n; i++) {
    while (top && st[top].val <= a[i]) top--;
    if (top) l[i] = st[top].pos + 1;
    st[++top] = Node(i, a[i]);
  }
  top = 0;
  for (int i = n; i >= 2; i--) {
    while (top && st[top].val < a[i]) top--;
    if (top)
      r[i] = st[top].pos - 1;
    else
      r[i] = n;
    st[++top] = Node(i, a[i]);
  }
  for (int i = 1; i <= q; i++) query[i].l = read(), query[i].r = read();
  for (int i = 1, left, right; i <= q; i++) {
    top = 0;
    for (int pos = query[i].l + 1; pos <= query[i].r; pos++) {
      left = max(query[i].l + 1, l[pos]), right = min(query[i].r, r[pos]);
      ans[i] += (long long)(right - pos + 1) * (pos - left + 1) * a[pos];
    }
  }
  for (int i = 1; i <= q; i++) printf("%I64d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int maxn = 100000 + 10;
struct Node {
  int pos, val;
  Node() {}
  Node(int pos, int val) : pos(pos), val(val) {}
} st[maxn];
struct Query {
  int l, r;
  bool operator<(const Query& tmp) const { return r < tmp.r; }
} query[maxn];
int n, q, top;
int a[maxn], l[maxn], r[maxn];
long long ans[maxn];
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
int main() {
  if (fopen("in.in", "r") != NULL) {
    freopen("in.in", "r", stdin);
    freopen("out.out", "w", stdout);
  }
  n = read(), q = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = n; i >= 1; i--) a[i] = abs(a[i] - a[i - 1]);
  for (int i = 2; i <= n; i++) {
    while (top && st[top].val <= a[i]) top--;
    if (top) l[i] = st[top].pos + 1;
    st[++top] = Node(i, a[i]);
  }
  top = 0;
  for (int i = n; i >= 2; i--) {
    while (top && st[top].val < a[i]) top--;
    if (top)
      r[i] = st[top].pos - 1;
    else
      r[i] = n;
    st[++top] = Node(i, a[i]);
  }
  for (int i = 1; i <= q; i++) query[i].l = read(), query[i].r = read();
  for (int i = 1, left, right; i <= q; i++) {
    top = 0;
    for (int pos = query[i].l + 1; pos <= query[i].r; pos++) {
      left = max(query[i].l + 1, l[pos]), right = min(query[i].r, r[pos]);
      ans[i] += (long long)(right - pos + 1) * (pos - left + 1) * a[pos];
    }
  }
  for (int i = 1; i <= q; i++) printf("%I64d\n", ans[i]);
  return 0;
}

```