#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
long long a[N], dis[N], tree[N << 2];
int up[N], down[N];
void update(int p, int l, int r, int x, long long y) {
  if (l == r) {
    tree[p] = max(tree[p], y);
    return;
  }
  int m = r + l >> 1;
  if (x <= m)
    update((p << 1), l, m, x, y);
  else
    update(((p << 1) | 1), m + 1, r, x, y);
  tree[p] = max(tree[(p << 1)], tree[((p << 1) | 1)]);
}
long long get(int p, int l, int r, int s, int t) {
  if (s > t) return 0;
  if (s <= l && r <= t) {
    return tree[p];
  }
  int m = l + r >> 1;
  if (t <= m) return get((p << 1), l, m, s, t);
  if (s > m) return get(((p << 1) | 1), m + 1, r, s, t);
  return max(get((p << 1), l, m, s, t), get(((p << 1) | 1), m + 1, r, s, t));
}
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &a[i]);
    }
    for (int i = 1; i < n; i++) {
      dis[i] = abs(a[i + 1] - a[i]);
    }
    memset(tree, 0, sizeof(tree));
    for (int i = 1; i < n; i++) {
      update(1, 1, n - 1, i, dis[i]);
    }
    int l, r, m;
    for (int i = 1; i < n; i++) {
      l = 1;
      r = i;
      while (l < r) {
        m = r + l >> 1;
        if (get(1, 1, n - 1, m, i - 1) >= dis[i])
          l = m + 1;
        else
          r = m;
      }
      down[i] = l;
      l = i;
      r = n - 1;
      while (l < r) {
        m = (r + l >> 1) + 1;
        if (get(1, 1, n - 1, i + 1, m) > dis[i])
          r = m - 1;
        else
          l = m;
      }
      up[i] = l;
    }
    for (int i = 0; i < q; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      long long ans = 0;
      for (int j = x; j < y; j++) {
        ans += (long long)(j - max(x, down[j]) + 1LL) *
               (min(y - 1, up[j]) - j + 1LL) * dis[j];
      }
      printf("%I64d\n", ans);
    }
  }
  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;
const int N = 1e5 + 100;
long long a[N], dis[N], tree[N << 2];
int up[N], down[N];
void update(int p, int l, int r, int x, long long y) {
  if (l == r) {
    tree[p] = max(tree[p], y);
    return;
  }
  int m = r + l >> 1;
  if (x <= m)
    update((p << 1), l, m, x, y);
  else
    update(((p << 1) | 1), m + 1, r, x, y);
  tree[p] = max(tree[(p << 1)], tree[((p << 1) | 1)]);
}
long long get(int p, int l, int r, int s, int t) {
  if (s > t) return 0;
  if (s <= l && r <= t) {
    return tree[p];
  }
  int m = l + r >> 1;
  if (t <= m) return get((p << 1), l, m, s, t);
  if (s > m) return get(((p << 1) | 1), m + 1, r, s, t);
  return max(get((p << 1), l, m, s, t), get(((p << 1) | 1), m + 1, r, s, t));
}
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &a[i]);
    }
    for (int i = 1; i < n; i++) {
      dis[i] = abs(a[i + 1] - a[i]);
    }
    memset(tree, 0, sizeof(tree));
    for (int i = 1; i < n; i++) {
      update(1, 1, n - 1, i, dis[i]);
    }
    int l, r, m;
    for (int i = 1; i < n; i++) {
      l = 1;
      r = i;
      while (l < r) {
        m = r + l >> 1;
        if (get(1, 1, n - 1, m, i - 1) >= dis[i])
          l = m + 1;
        else
          r = m;
      }
      down[i] = l;
      l = i;
      r = n - 1;
      while (l < r) {
        m = (r + l >> 1) + 1;
        if (get(1, 1, n - 1, i + 1, m) > dis[i])
          r = m - 1;
        else
          l = m;
      }
      up[i] = l;
    }
    for (int i = 0; i < q; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      long long ans = 0;
      for (int j = x; j < y; j++) {
        ans += (long long)(j - max(x, down[j]) + 1LL) *
               (min(y - 1, up[j]) - j + 1LL) * dis[j];
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

```