#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 10;
const long long MAXL = 20;
long long n, m;
long long v[MAXN], s[MAXN];
long long rmq[MAXL][MAXN];
long long resp[MAXN];
long long best[MAXN];
void buildRMQ() {
  for (long long k = 0; k < n; k++) rmq[0][k] = s[k];
  for (long long i = 1; i < MAXL; i++)
    for (long long k = 0; k < MAXN; k++)
      if (k + (1 << i) <= n)
        rmq[i][k] = max(rmq[i - 1][k], rmq[i - 1][k + (1 << i - 1)]);
}
long long RMQ(long long a, long long b) {
  long long size = b - a + 1;
  long long log = 31 - __builtin_clz(size);
  return max(rmq[log][a], rmq[log][b - (1 << log) + 1]);
}
long long bs(long long ini, long long fim, long long pos) {
  while (ini < fim) {
    long long mid = (ini + fim + 1) / 2;
    if (RMQ(mid, pos - 1) >= s[pos])
      ini = mid;
    else
      fim = mid - 1;
  }
  return ini;
}
int main() {
  scanf("%lld %lld", &n, &m);
  for (long long i = 0; i < n; i++) scanf("%lld", &v[i]);
  s[0] = 0;
  for (long long i = 1; i < n; i++) s[i] = abs(v[i] - v[i - 1]);
  buildRMQ();
  while (m--) {
    long long p, q;
    scanf("%lld %lld", &p, &q);
    p--;
    q--;
    long long ans = 0;
    stack<pair<long long, long long> > st;
    for (long long i = p; i <= q; i++) {
      while (st.size() && st.top().first <= s[i]) st.pop();
      if (st.size())
        best[i] = st.top().second;
      else
        best[i] = 0;
      st.push(pair<long long, long long>(s[i], i));
    }
    for (long long i = p; i <= q; i++) {
      resp[i] = 0;
      long long bt = best[i];
      if (s[bt] >= s[i]) {
        resp[i] = s[i] * (i - bt) + resp[bt];
      } else {
        resp[i] = s[i] * (i - p);
      }
      ans += resp[i];
    }
    printf("%lld\n", ans);
  }
}

### 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>
using namespace std;
const long long MAXN = 1e5 + 10;
const long long MAXL = 20;
long long n, m;
long long v[MAXN], s[MAXN];
long long rmq[MAXL][MAXN];
long long resp[MAXN];
long long best[MAXN];
void buildRMQ() {
  for (long long k = 0; k < n; k++) rmq[0][k] = s[k];
  for (long long i = 1; i < MAXL; i++)
    for (long long k = 0; k < MAXN; k++)
      if (k + (1 << i) <= n)
        rmq[i][k] = max(rmq[i - 1][k], rmq[i - 1][k + (1 << i - 1)]);
}
long long RMQ(long long a, long long b) {
  long long size = b - a + 1;
  long long log = 31 - __builtin_clz(size);
  return max(rmq[log][a], rmq[log][b - (1 << log) + 1]);
}
long long bs(long long ini, long long fim, long long pos) {
  while (ini < fim) {
    long long mid = (ini + fim + 1) / 2;
    if (RMQ(mid, pos - 1) >= s[pos])
      ini = mid;
    else
      fim = mid - 1;
  }
  return ini;
}
int main() {
  scanf("%lld %lld", &n, &m);
  for (long long i = 0; i < n; i++) scanf("%lld", &v[i]);
  s[0] = 0;
  for (long long i = 1; i < n; i++) s[i] = abs(v[i] - v[i - 1]);
  buildRMQ();
  while (m--) {
    long long p, q;
    scanf("%lld %lld", &p, &q);
    p--;
    q--;
    long long ans = 0;
    stack<pair<long long, long long> > st;
    for (long long i = p; i <= q; i++) {
      while (st.size() && st.top().first <= s[i]) st.pop();
      if (st.size())
        best[i] = st.top().second;
      else
        best[i] = 0;
      st.push(pair<long long, long long>(s[i], i));
    }
    for (long long i = p; i <= q; i++) {
      resp[i] = 0;
      long long bt = best[i];
      if (s[bt] >= s[i]) {
        resp[i] = s[i] * (i - bt) + resp[bt];
      } else {
        resp[i] = s[i] * (i - p);
      }
      ans += resp[i];
    }
    printf("%lld\n", ans);
  }
}

```