#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
void qread(int &x) {
  int neg = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
const int maxn = 100005;
const int mlog = 20;
int n, q;
long long ans;
int a[maxn], l2[maxn];
pair<int, int> mx[maxn][mlog];
pair<int, int> getval(int l, int r) {
  int len = r - l + 1, pw = l2[len];
  return max(mx[l][pw], mx[r - (1 << pw) + 1][pw]);
}
void solve(int l, int r) {
  if (l == r) return;
  pair<int, int> cur = getval(l, r - 1);
  ans += 1LL * (cur.second - l + 1) * (r - (cur.second + 1) + 1) * cur.first;
  solve(l, cur.second);
  solve(cur.second + 1, r);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int(i) = 0; (i) < n; i++) scanf("%d", &a[i]);
  for (int(i) = 0; (i) < n - 1; i++)
    mx[i][0] = make_pair(abs(a[i + 1] - a[i]), i);
  for (int(j) = 1; (j) <= mlog - 1; j++)
    for (int(i) = 0; (i) < n - 1; i++) {
      if (i + (1 << j) - 1 >= n)
        mx[i][j] = make_pair(-1, -1);
      else
        mx[i][j] = max(mx[i][j - 1], mx[i + (1 << j - 1)][j - 1]);
    }
  l2[1] = 0;
  for (int i = 2; i <= n; i++) {
    l2[i] = l2[i - 1];
    if (i == (1 << l2[i] + 1)) l2[i]++;
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    ans = 0;
    solve(l, r);
    printf("%I64d\n", ans);
  }
  return 0;
}

### 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>
#pragma GCC optimize(2)
using namespace std;
void qread(int &x) {
  int neg = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
const int maxn = 100005;
const int mlog = 20;
int n, q;
long long ans;
int a[maxn], l2[maxn];
pair<int, int> mx[maxn][mlog];
pair<int, int> getval(int l, int r) {
  int len = r - l + 1, pw = l2[len];
  return max(mx[l][pw], mx[r - (1 << pw) + 1][pw]);
}
void solve(int l, int r) {
  if (l == r) return;
  pair<int, int> cur = getval(l, r - 1);
  ans += 1LL * (cur.second - l + 1) * (r - (cur.second + 1) + 1) * cur.first;
  solve(l, cur.second);
  solve(cur.second + 1, r);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int(i) = 0; (i) < n; i++) scanf("%d", &a[i]);
  for (int(i) = 0; (i) < n - 1; i++)
    mx[i][0] = make_pair(abs(a[i + 1] - a[i]), i);
  for (int(j) = 1; (j) <= mlog - 1; j++)
    for (int(i) = 0; (i) < n - 1; i++) {
      if (i + (1 << j) - 1 >= n)
        mx[i][j] = make_pair(-1, -1);
      else
        mx[i][j] = max(mx[i][j - 1], mx[i + (1 << j - 1)][j - 1]);
    }
  l2[1] = 0;
  for (int i = 2; i <= n; i++) {
    l2[i] = l2[i - 1];
    if (i == (1 << l2[i] + 1)) l2[i]++;
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    ans = 0;
    solve(l, r);
    printf("%I64d\n", ans);
  }
  return 0;
}

```