#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = 2 * acos(0);
const int maxn = 100010;
int n, q;
int a[maxn];
int d[maxn];
int l[maxn];
int r[maxn];
void solve() {
  scanf("%d%d", &n, &q);
  for (int i = (0); i < (n); i++) scanf("%d", a + i);
  for (int i = (0); i < (n - 1); i++) d[i] = abs(a[i + 1] - a[i]);
  for (int i = (0); i < (n - 1); i++) l[i] = r[i] = i;
  for (int i = (1); i < (n - 1); i++) {
    int cur = i;
    while (cur > 0 && d[cur - 1] <= d[i]) cur = l[cur - 1];
    l[i] = cur;
  }
  for (int i = (n - 2) - 1; i >= (0); i--) {
    int cur = i;
    while (cur < n - 2 && d[cur + 1] < d[i]) cur = r[cur + 1];
    r[i] = cur;
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y -= 2;
    long long ans = 0;
    for (int i = (x); i < (y + 1); i++) {
      int u = i - max(x, l[i]);
      int v = min(y, r[i]) - i;
      if (u >= 0 && v >= 0) ans += 1LL * d[i] * (u + 1) * (v + 1);
    }
    printf("%I64d\n", ans);
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = 2 * acos(0);
const int maxn = 100010;
int n, q;
int a[maxn];
int d[maxn];
int l[maxn];
int r[maxn];
void solve() {
  scanf("%d%d", &n, &q);
  for (int i = (0); i < (n); i++) scanf("%d", a + i);
  for (int i = (0); i < (n - 1); i++) d[i] = abs(a[i + 1] - a[i]);
  for (int i = (0); i < (n - 1); i++) l[i] = r[i] = i;
  for (int i = (1); i < (n - 1); i++) {
    int cur = i;
    while (cur > 0 && d[cur - 1] <= d[i]) cur = l[cur - 1];
    l[i] = cur;
  }
  for (int i = (n - 2) - 1; i >= (0); i--) {
    int cur = i;
    while (cur < n - 2 && d[cur + 1] < d[i]) cur = r[cur + 1];
    r[i] = cur;
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y -= 2;
    long long ans = 0;
    for (int i = (x); i < (y + 1); i++) {
      int u = i - max(x, l[i]);
      int v = min(y, r[i]) - i;
      if (u >= 0 && v >= 0) ans += 1LL * d[i] * (u + 1) * (v + 1);
    }
    printf("%I64d\n", ans);
  }
}
int main() {
  solve();
  return 0;
}

```