#include <bits/stdc++.h>
using namespace std;
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
const double EPS = 1e-9;
const double PI = acos(-1.);
const int INF = 1e9;
const int MAXN = 1e5 + 5, MAXLOG = 17;
const long long MOD = 1e9 + 7;
int a[MAXN], h[MAXN], dif_ind[MAXN];
pair<int, int> tt[MAXN][MAXLOG];
void build(int n) {
  int i, j, st2;
  for ((i) = (0); (i) < (int)(n); (i)++) {
    tt[i][0] = make_pair(a[i], i);
  }
  for (j = 1, st2 = 2; j < MAXLOG; j++, st2 <<= 1) {
    int offset = st2 >> 1;
    for (i = st2; i < min(MAXN, (st2 << 1)); i++) {
      dif_ind[i] = j;
    }
    for (i = 0; i < n; i++) {
      tt[i][j] = max(tt[i][j - 1], tt[min(i + offset, n - 1)][j - 1]);
    }
  }
}
pair<int, int> get_val(int l, int r) {
  int ind = dif_ind[r - l];
  return max(tt[l][ind], tt[r - (1 << ind) + 1][ind]);
}
long long req_sol(int n, int l, int r) {
  if (l > r) {
    return 0ll;
  }
  pair<int, int> pp = get_val(l, r);
  int mid = pp.second;
  long long v0 =
      (long long)pp.first * (long long)(mid - l + 1) * (long long)(r - mid + 1);
  long long v1 = req_sol(n, l, mid - 1);
  long long v2 = req_sol(n, mid + 1, r);
  return v0 + v1 + v2;
}
void solve() {
  int n, i, j, q, l, r;
  scanf("%d%d", &n, &q);
  for ((i) = (0); (i) < (int)(n); (i)++) {
    scanf("%d", &h[i]);
    if (i) {
      a[i - 1] = abs(h[i] - h[i - 1]);
    }
  }
  n--;
  build(n);
  for ((i) = (0); (i) < (int)(q); (i)++) {
    scanf("%d%d", &l, &r);
    long long ans = req_sol(n, l - 1, r - 2);
    printf("%lld\n", ans);
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}

### 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 T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
const double EPS = 1e-9;
const double PI = acos(-1.);
const int INF = 1e9;
const int MAXN = 1e5 + 5, MAXLOG = 17;
const long long MOD = 1e9 + 7;
int a[MAXN], h[MAXN], dif_ind[MAXN];
pair<int, int> tt[MAXN][MAXLOG];
void build(int n) {
  int i, j, st2;
  for ((i) = (0); (i) < (int)(n); (i)++) {
    tt[i][0] = make_pair(a[i], i);
  }
  for (j = 1, st2 = 2; j < MAXLOG; j++, st2 <<= 1) {
    int offset = st2 >> 1;
    for (i = st2; i < min(MAXN, (st2 << 1)); i++) {
      dif_ind[i] = j;
    }
    for (i = 0; i < n; i++) {
      tt[i][j] = max(tt[i][j - 1], tt[min(i + offset, n - 1)][j - 1]);
    }
  }
}
pair<int, int> get_val(int l, int r) {
  int ind = dif_ind[r - l];
  return max(tt[l][ind], tt[r - (1 << ind) + 1][ind]);
}
long long req_sol(int n, int l, int r) {
  if (l > r) {
    return 0ll;
  }
  pair<int, int> pp = get_val(l, r);
  int mid = pp.second;
  long long v0 =
      (long long)pp.first * (long long)(mid - l + 1) * (long long)(r - mid + 1);
  long long v1 = req_sol(n, l, mid - 1);
  long long v2 = req_sol(n, mid + 1, r);
  return v0 + v1 + v2;
}
void solve() {
  int n, i, j, q, l, r;
  scanf("%d%d", &n, &q);
  for ((i) = (0); (i) < (int)(n); (i)++) {
    scanf("%d", &h[i]);
    if (i) {
      a[i - 1] = abs(h[i] - h[i - 1]);
    }
  }
  n--;
  build(n);
  for ((i) = (0); (i) < (int)(q); (i)++) {
    scanf("%d%d", &l, &r);
    long long ans = req_sol(n, l - 1, r - 2);
    printf("%lld\n", ans);
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}

```