#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
template <class T>
inline void gmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void gmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
const int N = 1e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647,
          ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567,
          ms127 = 2139062143;
const double PI = acos(-1.0), eps = 1e-8;
void fre() {
  freopen("/Users/luras/Desktop/in.txt", "r", stdin);
  freopen("/Users/luras/Desktop/out.txt", "w", stdout);
}
const int INF = 1e9;
int casenum, casei;
struct A {
  int l, r, max;
} a[1 << 18];
int n, m, l, r;
int vv[N], v[N], L[N], R[N];
void build(int o, int l, int r) {
  a[o].l = l;
  a[o].r = r;
  if (l == r) {
    a[o].max = v[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  a[o].max = max(a[o << 1].max, a[o << 1 | 1].max);
}
int V;
int trylft(int o, int l, int r) {
  if (a[o].max < V)
    return l;
  else if (a[o].l == a[o].r)
    return r + 1;
  int mid = (a[o].l + a[o].r) >> 1;
  if (a[o].l == l && a[o].r == r) {
    if (a[o << 1 | 1].max < V)
      return trylft(o << 1, l, mid);
    else
      return trylft(o << 1 | 1, mid + 1, r);
  }
  if (r <= mid)
    return trylft(o << 1, l, r);
  else if (l > mid)
    return trylft(o << 1 | 1, l, r);
  else {
    int pos = trylft(o << 1 | 1, mid + 1, r);
    if (pos > mid + 1)
      return pos;
    else
      return trylft(o << 1, l, mid);
  }
}
int tryrgt(int o, int l, int r) {
  if (a[o].max <= V)
    return r;
  else if (a[o].l == a[o].r)
    return l - 1;
  int mid = (a[o].l + a[o].r) >> 1;
  if (a[o].l == l && a[o].r == r) {
    if (a[o << 1].max <= V)
      return tryrgt(o << 1 | 1, mid + 1, r);
    else
      return tryrgt(o << 1, l, mid);
  }
  if (r <= mid)
    return tryrgt(o << 1, l, r);
  else if (l > mid)
    return tryrgt(o << 1 | 1, l, r);
  else {
    int pos = tryrgt(o << 1, l, mid);
    if (pos < mid)
      return pos;
    else
      return tryrgt(o << 1 | 1, mid + 1, r);
  }
}
void init() {
  for (int j = 1; j <= n; j++) {
    V = v[j];
    if (j == 1)
      L[j] = 1;
    else
      L[j] = trylft(1, 1, j - 1);
    if (j == n)
      R[j] = n;
    else
      R[j] = tryrgt(1, j + 1, n);
  }
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= n; i++) scanf("%d", &vv[i]);
    n--;
    for (int i = 1; i <= n; i++) v[i] = fabs(vv[i + 1] - vv[i]);
    build(1, 1, n);
    init();
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &l, &r);
      r--;
      long long ans = 0;
      for (int j = l; j <= r; j++) {
        ans +=
            (long long)(j - max(L[j], l) + 1) * (min(R[j], r) - j + 1) * v[j];
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:102400000,102400000")
template <class T>
inline void gmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void gmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
const int N = 1e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647,
          ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567,
          ms127 = 2139062143;
const double PI = acos(-1.0), eps = 1e-8;
void fre() {
  freopen("/Users/luras/Desktop/in.txt", "r", stdin);
  freopen("/Users/luras/Desktop/out.txt", "w", stdout);
}
const int INF = 1e9;
int casenum, casei;
struct A {
  int l, r, max;
} a[1 << 18];
int n, m, l, r;
int vv[N], v[N], L[N], R[N];
void build(int o, int l, int r) {
  a[o].l = l;
  a[o].r = r;
  if (l == r) {
    a[o].max = v[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  a[o].max = max(a[o << 1].max, a[o << 1 | 1].max);
}
int V;
int trylft(int o, int l, int r) {
  if (a[o].max < V)
    return l;
  else if (a[o].l == a[o].r)
    return r + 1;
  int mid = (a[o].l + a[o].r) >> 1;
  if (a[o].l == l && a[o].r == r) {
    if (a[o << 1 | 1].max < V)
      return trylft(o << 1, l, mid);
    else
      return trylft(o << 1 | 1, mid + 1, r);
  }
  if (r <= mid)
    return trylft(o << 1, l, r);
  else if (l > mid)
    return trylft(o << 1 | 1, l, r);
  else {
    int pos = trylft(o << 1 | 1, mid + 1, r);
    if (pos > mid + 1)
      return pos;
    else
      return trylft(o << 1, l, mid);
  }
}
int tryrgt(int o, int l, int r) {
  if (a[o].max <= V)
    return r;
  else if (a[o].l == a[o].r)
    return l - 1;
  int mid = (a[o].l + a[o].r) >> 1;
  if (a[o].l == l && a[o].r == r) {
    if (a[o << 1].max <= V)
      return tryrgt(o << 1 | 1, mid + 1, r);
    else
      return tryrgt(o << 1, l, mid);
  }
  if (r <= mid)
    return tryrgt(o << 1, l, r);
  else if (l > mid)
    return tryrgt(o << 1 | 1, l, r);
  else {
    int pos = tryrgt(o << 1, l, mid);
    if (pos < mid)
      return pos;
    else
      return tryrgt(o << 1 | 1, mid + 1, r);
  }
}
void init() {
  for (int j = 1; j <= n; j++) {
    V = v[j];
    if (j == 1)
      L[j] = 1;
    else
      L[j] = trylft(1, 1, j - 1);
    if (j == n)
      R[j] = n;
    else
      R[j] = tryrgt(1, j + 1, n);
  }
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= n; i++) scanf("%d", &vv[i]);
    n--;
    for (int i = 1; i <= n; i++) v[i] = fabs(vv[i + 1] - vv[i]);
    build(1, 1, n);
    init();
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &l, &r);
      r--;
      long long ans = 0;
      for (int j = l; j <= r; j++) {
        ans +=
            (long long)(j - max(L[j], l) + 1) * (min(R[j], r) - j + 1) * v[j];
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```