#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int N = 201000;
int stk[N], top, a[N], l[N], r[N], b[N], vis[N];
int n, q, pl, pr;
long long ans;
void dfs(int u, int *a, int pl, int pr) {
  if (l[u]) dfs(l[u], a, pl, u - 1);
  if (r[u]) dfs(r[u], a, u + 1, pr);
  ans += (long long)-a[u] * (pr - u + 1) * (u - pl + 1);
}
void build(int *a, int n) {
  int top = 0;
  for (int i = 1; i < n + 1; i++) l[i] = r[i] = 0, vis[i] = 0;
  for (int i = 1; i < n + 1; i++) {
    int k = top;
    while (k > 0 && a[stk[k - 1]] > a[i]) --k;
    if (k) r[stk[k - 1]] = i;
    if (k < top) l[i] = stk[k];
    stk[k++] = i;
    top = k;
  }
  for (int i = 1; i < n + 1; i++) vis[l[i]] = vis[r[i]] = 1;
  int rt = 0;
  for (int i = 1; i < n + 1; i++)
    if (!vis[i]) rt = i;
  ans = 0;
  dfs(rt, a, 1, n);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n + 1; i++) scanf("%d", a + i);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &pl, &pr);
    for (int j = pl; j < pr; j++) b[j - pl + 1] = -abs(a[j + 1] - a[j]);
    build(b, pr - pl);
    printf("%lld\n", ans);
  }
}

### Prompt

Your challenge is to write 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 mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int N = 201000;
int stk[N], top, a[N], l[N], r[N], b[N], vis[N];
int n, q, pl, pr;
long long ans;
void dfs(int u, int *a, int pl, int pr) {
  if (l[u]) dfs(l[u], a, pl, u - 1);
  if (r[u]) dfs(r[u], a, u + 1, pr);
  ans += (long long)-a[u] * (pr - u + 1) * (u - pl + 1);
}
void build(int *a, int n) {
  int top = 0;
  for (int i = 1; i < n + 1; i++) l[i] = r[i] = 0, vis[i] = 0;
  for (int i = 1; i < n + 1; i++) {
    int k = top;
    while (k > 0 && a[stk[k - 1]] > a[i]) --k;
    if (k) r[stk[k - 1]] = i;
    if (k < top) l[i] = stk[k];
    stk[k++] = i;
    top = k;
  }
  for (int i = 1; i < n + 1; i++) vis[l[i]] = vis[r[i]] = 1;
  int rt = 0;
  for (int i = 1; i < n + 1; i++)
    if (!vis[i]) rt = i;
  ans = 0;
  dfs(rt, a, 1, n);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n + 1; i++) scanf("%d", a + i);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &pl, &pr);
    for (int j = pl; j < pr; j++) b[j - pl + 1] = -abs(a[j + 1] - a[j]);
    build(b, pr - pl);
    printf("%lld\n", ans);
  }
}

```