#include <bits/stdc++.h>
using namespace std;
template <typename T>
T nextInt() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * p;
}
const int maxN = (int)1e5 + 10;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
int a[maxN];
int leftest[maxN];
int rightest[maxN];
pair<int, int> v[maxN];
int main() {
  srand(time(NULL));
  int n, m;
  n = nextInt<int>();
  m = nextInt<int>();
  for (int i = 1; i <= n; ++i) {
    a[i] = nextInt<int>();
  }
  for (int i = 1; i < n; ++i) {
    v[i] = make_pair(abs(a[i] - a[i + 1]), i);
  }
  sort(v + 1, v + n);
  reverse(v + 1, v + n);
  set<int> avail;
  avail.insert(0);
  avail.insert(n);
  for (int i = 1; i < n; ++i) {
    auto iter = avail.lower_bound(v[i].second);
    leftest[i] = *--iter;
    iter = avail.lower_bound(v[i].second);
    rightest[i] = *iter;
    avail.insert(v[i].second);
  }
  while (m--) {
    int l = nextInt<int>();
    int r = nextInt<int>();
    long long answer = 0LL;
    set<int> avail;
    avail.insert(0);
    avail.insert(r - l + 1);
    for (int i = 1; i < n; ++i) {
      if (l <= v[i].second && v[i].second < r) {
        int L = max(l - 1, leftest[i]);
        int R = min(r, rightest[i]);
        answer += v[i].first * 1LL * (v[i].second - L) * (R - v[i].second);
      }
    }
    printf("%I64d\n", answer);
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <typename T>
T nextInt() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * p;
}
const int maxN = (int)1e5 + 10;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
int a[maxN];
int leftest[maxN];
int rightest[maxN];
pair<int, int> v[maxN];
int main() {
  srand(time(NULL));
  int n, m;
  n = nextInt<int>();
  m = nextInt<int>();
  for (int i = 1; i <= n; ++i) {
    a[i] = nextInt<int>();
  }
  for (int i = 1; i < n; ++i) {
    v[i] = make_pair(abs(a[i] - a[i + 1]), i);
  }
  sort(v + 1, v + n);
  reverse(v + 1, v + n);
  set<int> avail;
  avail.insert(0);
  avail.insert(n);
  for (int i = 1; i < n; ++i) {
    auto iter = avail.lower_bound(v[i].second);
    leftest[i] = *--iter;
    iter = avail.lower_bound(v[i].second);
    rightest[i] = *iter;
    avail.insert(v[i].second);
  }
  while (m--) {
    int l = nextInt<int>();
    int r = nextInt<int>();
    long long answer = 0LL;
    set<int> avail;
    avail.insert(0);
    avail.insert(r - l + 1);
    for (int i = 1; i < n; ++i) {
      if (l <= v[i].second && v[i].second < r) {
        int L = max(l - 1, leftest[i]);
        int R = min(r, rightest[i]);
        answer += v[i].first * 1LL * (v[i].second - L) * (R - v[i].second);
      }
    }
    printf("%I64d\n", answer);
  }
  return 0;
}

```