#include <bits/stdc++.h>
using namespace std;
int n, q;
long long a[100001];
int L[100];
int R[100];
set<int> bds;
long long low[100001];
long long high[100001];
struct slope {
  slope(long long val, long long loc) : val(val), loc(loc) {}
  long long val;
  int loc;
};
struct slcmp {
  bool operator()(slope &a, slope &b) { return a.val < b.val; }
};
priority_queue<slope, vector<slope>, slcmp> pq;
void solve_testcase() {
  bds.clear();
  bds.insert(-1);
  bds.insert(n);
  for (int loc = 1; loc < n; loc++) {
    long long val = a[loc] - a[loc + 1];
    if (val < 0) val *= -1;
    pq.push(slope(val, loc));
  }
  while (!pq.empty()) {
    slope s = pq.top();
    pq.pop();
    set<int>::iterator it = bds.lower_bound(s.loc);
    high[s.loc] = *it;
    it--;
    low[s.loc] = *it;
    bds.insert(s.loc);
  }
  for (int i = 0; i < q; i++) {
    int l = L[i];
    int r = R[i];
    long long res = 0;
    for (int loc = l; loc < r; loc++) {
      long long lo = max(low[loc], (long long)l - 1);
      long long hi = min(high[loc], (long long)r);
      long long mult = ((long long)loc - lo) * (hi - (long long)loc);
      long long val = a[loc] - a[loc + 1];
      if (val < 0) val *= -1;
      res += mult * val;
    }
    cout << res << endl;
  }
}
bool read_testcase() {
  if (scanf("%d %d", &n, &q) != 2) return false;
  for (int i = 1; i <= n; i++) {
    int aa;
    scanf("%d", &aa);
    a[i] = aa;
  }
  for (int i = 0; i < q; i++) scanf("%d %d", L + i, R + i);
  return true;
}
int main(int argc, char **argv) {
  if (argc > 1 && argv[1][0] != 'x') freopen(argv[1], "r", stdin);
  if (argc > 2) freopen(argv[2], "w", stdout);
  int T = 1;
  clock_t cur, prev = clock();
  while (read_testcase()) {
    fprintf(stderr, ">>>Solving testcase %d<<<<\n", T);
    solve_testcase();
    cur = clock();
    int dur = (cur - prev) * 1000 / CLOCKS_PER_SEC;
    fprintf(stderr, "____Case %d done in %dms.\n", T, dur);
    prev = cur;
    T++;
  }
}

### Prompt

In Cpp, your task is to solve 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;
int n, q;
long long a[100001];
int L[100];
int R[100];
set<int> bds;
long long low[100001];
long long high[100001];
struct slope {
  slope(long long val, long long loc) : val(val), loc(loc) {}
  long long val;
  int loc;
};
struct slcmp {
  bool operator()(slope &a, slope &b) { return a.val < b.val; }
};
priority_queue<slope, vector<slope>, slcmp> pq;
void solve_testcase() {
  bds.clear();
  bds.insert(-1);
  bds.insert(n);
  for (int loc = 1; loc < n; loc++) {
    long long val = a[loc] - a[loc + 1];
    if (val < 0) val *= -1;
    pq.push(slope(val, loc));
  }
  while (!pq.empty()) {
    slope s = pq.top();
    pq.pop();
    set<int>::iterator it = bds.lower_bound(s.loc);
    high[s.loc] = *it;
    it--;
    low[s.loc] = *it;
    bds.insert(s.loc);
  }
  for (int i = 0; i < q; i++) {
    int l = L[i];
    int r = R[i];
    long long res = 0;
    for (int loc = l; loc < r; loc++) {
      long long lo = max(low[loc], (long long)l - 1);
      long long hi = min(high[loc], (long long)r);
      long long mult = ((long long)loc - lo) * (hi - (long long)loc);
      long long val = a[loc] - a[loc + 1];
      if (val < 0) val *= -1;
      res += mult * val;
    }
    cout << res << endl;
  }
}
bool read_testcase() {
  if (scanf("%d %d", &n, &q) != 2) return false;
  for (int i = 1; i <= n; i++) {
    int aa;
    scanf("%d", &aa);
    a[i] = aa;
  }
  for (int i = 0; i < q; i++) scanf("%d %d", L + i, R + i);
  return true;
}
int main(int argc, char **argv) {
  if (argc > 1 && argv[1][0] != 'x') freopen(argv[1], "r", stdin);
  if (argc > 2) freopen(argv[2], "w", stdout);
  int T = 1;
  clock_t cur, prev = clock();
  while (read_testcase()) {
    fprintf(stderr, ">>>Solving testcase %d<<<<\n", T);
    solve_testcase();
    cur = clock();
    int dur = (cur - prev) * 1000 / CLOCKS_PER_SEC;
    fprintf(stderr, "____Case %d done in %dms.\n", T, dur);
    prev = cur;
    T++;
  }
}

```