#include <bits/stdc++.h>
using namespace std;
const int MN = 100111;
int n, a[MN], l[MN], r[MN], st[MN], top;
long long x[MN];
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout << (fixed) << setprecision(9);
  int q;
  while (scanf("%d%d", &n, &q) == 2) {
    for (int i = (1), _b = (n); i <= _b; ++i) scanf("%d", &a[i]);
    while (q--) {
      int u, v;
      scanf("%d%d", &u, &v);
      int sx = 0;
      for (int i = (u), _b = (v - 1); i <= _b; ++i)
        x[++sx] = abs(a[i + 1] - a[i]);
      st[0] = 0;
      top = 0;
      for (int i = (1), _b = (sx); i <= _b; ++i) {
        while (top && x[st[top]] <= x[i]) --top;
        l[i] = st[top] + 1;
        st[++top] = i;
      }
      st[0] = sx + 1;
      top = 0;
      for (int i = (sx), _b = (1); i >= _b; --i) {
        while (top && x[st[top]] < x[i]) --top;
        r[i] = st[top] - 1;
        st[++top] = i;
      }
      long long res = 0;
      for (int i = (1), _b = (sx); i <= _b; ++i) {
        res += x[i] * (i - l[i] + 1LL) * (r[i] - i + 1LL);
      }
      printf("%lld\n", res);
    }
  }
}

### 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 int MN = 100111;
int n, a[MN], l[MN], r[MN], st[MN], top;
long long x[MN];
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout << (fixed) << setprecision(9);
  int q;
  while (scanf("%d%d", &n, &q) == 2) {
    for (int i = (1), _b = (n); i <= _b; ++i) scanf("%d", &a[i]);
    while (q--) {
      int u, v;
      scanf("%d%d", &u, &v);
      int sx = 0;
      for (int i = (u), _b = (v - 1); i <= _b; ++i)
        x[++sx] = abs(a[i + 1] - a[i]);
      st[0] = 0;
      top = 0;
      for (int i = (1), _b = (sx); i <= _b; ++i) {
        while (top && x[st[top]] <= x[i]) --top;
        l[i] = st[top] + 1;
        st[++top] = i;
      }
      st[0] = sx + 1;
      top = 0;
      for (int i = (sx), _b = (1); i >= _b; --i) {
        while (top && x[st[top]] < x[i]) --top;
        r[i] = st[top] - 1;
        st[++top] = i;
      }
      long long res = 0;
      for (int i = (1), _b = (sx); i <= _b; ++i) {
        res += x[i] * (i - l[i] + 1LL) * (r[i] - i + 1LL);
      }
      printf("%lld\n", res);
    }
  }
}

```