#include <bits/stdc++.h>
using namespace std;
const long long maxn = 100010;
long long n, q, cnt, h[maxn], S[maxn], top, L[maxn], R[maxn];
long long query(long long len) { return len * (len + 1) / 2; }
void solve(long long l, long long r) {
  long long res = 0;
  top = 0;
  memset(L, 0, sizeof(L));
  memset(R, 0, sizeof(R));
  memset(S, 0, sizeof(S));
  S[0] = l - 1;
  for (long long i = l; i <= r; i++) {
    while (top && h[S[top]] <= h[i]) top--;
    L[i] = S[top] + 1;
    S[++top] = i;
  }
  top = 0;
  memset(S, 0, sizeof(S));
  S[0] = r + 1;
  for (long long i = r; i >= l; i--) {
    while (top && h[S[top]] < h[i]) top--;
    R[i] = S[top] - 1;
    S[++top] = i;
  }
  for (long long i = l; i <= r; i++) {
    res += h[i] * ((i - L[i] + 1) * (R[i] - i + 1));
  }
  printf("%lld\n", res);
}
int main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &h[i]);
  for (long long i = n; i >= 2; i--) h[i] = abs(h[i] - h[i - 1]);
  while (q--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    if (l == r)
      puts("0");
    else {
      l++;
      solve(l, r);
    }
  }
  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 maxn = 100010;
long long n, q, cnt, h[maxn], S[maxn], top, L[maxn], R[maxn];
long long query(long long len) { return len * (len + 1) / 2; }
void solve(long long l, long long r) {
  long long res = 0;
  top = 0;
  memset(L, 0, sizeof(L));
  memset(R, 0, sizeof(R));
  memset(S, 0, sizeof(S));
  S[0] = l - 1;
  for (long long i = l; i <= r; i++) {
    while (top && h[S[top]] <= h[i]) top--;
    L[i] = S[top] + 1;
    S[++top] = i;
  }
  top = 0;
  memset(S, 0, sizeof(S));
  S[0] = r + 1;
  for (long long i = r; i >= l; i--) {
    while (top && h[S[top]] < h[i]) top--;
    R[i] = S[top] - 1;
    S[++top] = i;
  }
  for (long long i = l; i <= r; i++) {
    res += h[i] * ((i - L[i] + 1) * (R[i] - i + 1));
  }
  printf("%lld\n", res);
}
int main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &h[i]);
  for (long long i = n; i >= 2; i--) h[i] = abs(h[i] - h[i - 1]);
  while (q--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    if (l == r)
      puts("0");
    else {
      l++;
      solve(l, r);
    }
  }
  return 0;
}

```