#include <bits/stdc++.h>
using namespace std;
const long long inf = (long long)5e18;
const long long N = 120000;
long long n, a[N], m, b[N];
long long pre[N], nxt[N];
long long q[N], top = 0;
long long _abs(long long x) { return x < 0 ? -x : x; }
void solve() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (i > 1) b[i - 1] = _abs(a[i] - a[i - 1]);
  }
  b[0] = b[n] = inf;
  q[++top] = 0;
  for (long long i = 1; i <= n; i++) {
    while (b[q[top]] < b[i]) nxt[q[top]] = i, top--;
    pre[i] = q[top];
    q[++top] = i;
  }
  while (m--) {
    long long ans = 0;
    long long l, r;
    scanf("%lld%lld", &l, &r);
    r--;
    for (long long i = l; i <= r; i++) {
      long long L = max(l, pre[i] + 1);
      long long R = min(r, nxt[i] - 1);
      ans += (R - i + 1) * (i - L + 1) * b[i];
    }
    printf("%lld\n", ans);
  }
}
signed main() {
  long long TTT = 1;
  while (TTT--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 inf = (long long)5e18;
const long long N = 120000;
long long n, a[N], m, b[N];
long long pre[N], nxt[N];
long long q[N], top = 0;
long long _abs(long long x) { return x < 0 ? -x : x; }
void solve() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (i > 1) b[i - 1] = _abs(a[i] - a[i - 1]);
  }
  b[0] = b[n] = inf;
  q[++top] = 0;
  for (long long i = 1; i <= n; i++) {
    while (b[q[top]] < b[i]) nxt[q[top]] = i, top--;
    pre[i] = q[top];
    q[++top] = i;
  }
  while (m--) {
    long long ans = 0;
    long long l, r;
    scanf("%lld%lld", &l, &r);
    r--;
    for (long long i = l; i <= r; i++) {
      long long L = max(l, pre[i] + 1);
      long long R = min(r, nxt[i] - 1);
      ans += (R - i + 1) * (i - L + 1) * b[i];
    }
    printf("%lld\n", ans);
  }
}
signed main() {
  long long TTT = 1;
  while (TTT--) {
    solve();
  }
  return 0;
}

```