#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-7;
const long long SZ = 100010, SSZ = 120, APB = 52, one = 11;
const long long INF = 0x7f7f7f7f7f7f7f7f, mod = 1000000007;
long long n, qnum, src[SZ], diff[SZ];
struct nd {
  long long ll, rr, val, id;
  nd(long long a = 0, long long b = 0, long long c = 0, long long d = 0)
      : ll(a), rr(b), val(c), id(d) {}
};
nd arr[SZ];
void its(long long l1, long long r1, long long l2, long long r2, long long &u,
         long long &v) {
  u = max(l1, l2);
  v = min(r1, r2);
  if (u > v) u = v = -1;
}
void init() {
  cin >> n >> qnum;
  for (long long i = 1; i <= n; ++i) cin >> src[i];
  for (long long i = 1; i <= n - 1; ++i) {
    diff[i] = abs(src[i + 1] - src[i]);
  }
  deque<nd> dq;
  for (long long i = 1; i <= n - 1; ++i) {
    long long ll = i;
    for (; dq.size() && dq.back().val <= diff[i];) {
      nd tmp = dq.back();
      dq.pop_back();
      tmp.rr = i - 1;
      arr[tmp.id] = tmp;
      ll = min(ll, tmp.ll);
    }
    dq.push_back(nd(ll, i, diff[i], i));
  }
  for (; dq.size();) {
    nd tmp = dq.back();
    dq.pop_back();
    tmp.rr = n - 1;
    arr[tmp.id] = tmp;
  }
  for (long long i = 1; i <= n - 1; ++i) {
    ++arr[i].rr;
  }
  for (long long i = 1; i <= qnum; ++i) {
    long long ll, rr, res = 0, u, v;
    cin >> ll >> rr;
    for (long long j = 1; j <= n - 1; ++j) {
      its(ll, rr, arr[j].ll, arr[j].rr, u, v);
      if (u != -1) {
        long long num1 = j - u + 1;
        long long num2 = v - j;
        if (num1 < 0) num1 = 0;
        if (num2 < 0) num2 = 0;
        res += num1 * num2 * diff[j];
      }
    }
    cout << res << endl;
  }
}
void work() {}
int main() {
  std::ios::sync_with_stdio(0);
  int casenum;
  {
    init();
    work();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const double EPS = 1e-7;
const long long SZ = 100010, SSZ = 120, APB = 52, one = 11;
const long long INF = 0x7f7f7f7f7f7f7f7f, mod = 1000000007;
long long n, qnum, src[SZ], diff[SZ];
struct nd {
  long long ll, rr, val, id;
  nd(long long a = 0, long long b = 0, long long c = 0, long long d = 0)
      : ll(a), rr(b), val(c), id(d) {}
};
nd arr[SZ];
void its(long long l1, long long r1, long long l2, long long r2, long long &u,
         long long &v) {
  u = max(l1, l2);
  v = min(r1, r2);
  if (u > v) u = v = -1;
}
void init() {
  cin >> n >> qnum;
  for (long long i = 1; i <= n; ++i) cin >> src[i];
  for (long long i = 1; i <= n - 1; ++i) {
    diff[i] = abs(src[i + 1] - src[i]);
  }
  deque<nd> dq;
  for (long long i = 1; i <= n - 1; ++i) {
    long long ll = i;
    for (; dq.size() && dq.back().val <= diff[i];) {
      nd tmp = dq.back();
      dq.pop_back();
      tmp.rr = i - 1;
      arr[tmp.id] = tmp;
      ll = min(ll, tmp.ll);
    }
    dq.push_back(nd(ll, i, diff[i], i));
  }
  for (; dq.size();) {
    nd tmp = dq.back();
    dq.pop_back();
    tmp.rr = n - 1;
    arr[tmp.id] = tmp;
  }
  for (long long i = 1; i <= n - 1; ++i) {
    ++arr[i].rr;
  }
  for (long long i = 1; i <= qnum; ++i) {
    long long ll, rr, res = 0, u, v;
    cin >> ll >> rr;
    for (long long j = 1; j <= n - 1; ++j) {
      its(ll, rr, arr[j].ll, arr[j].rr, u, v);
      if (u != -1) {
        long long num1 = j - u + 1;
        long long num2 = v - j;
        if (num1 < 0) num1 = 0;
        if (num2 < 0) num2 = 0;
        res += num1 * num2 * diff[j];
      }
    }
    cout << res << endl;
  }
}
void work() {}
int main() {
  std::ios::sync_with_stdio(0);
  int casenum;
  {
    init();
    work();
  }
  return 0;
}

```