#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
const int MAXN = 500010;
long long s[MAXN], n, m, prox[MAXN], ant[MAXN];
stack<pair<long long, long long> > dir, esq;
int main() {
  cin >> n >> m;
  memset(ant, -1, sizeof(ant));
  for (int a = 0; a <= n; a++) prox[a] = n - 1;
  for (int a = 0; a < n; a++) cin >> s[a];
  for (int a = 0; a < n - 1; a++) {
    s[a] = abs(s[a + 1] - s[a]);
  }
  dir.push(make_pair(s[0], 0));
  pair<long long, long long> topo;
  for (int a = 1; a < n - 1; a++) {
    topo = dir.top();
    while (!dir.empty() && s[a] >= topo.first) {
      prox[topo.second] = a;
      dir.pop();
      if (!dir.empty()) topo = dir.top();
    }
    if (!dir.empty()) {
      topo = dir.top();
      ant[a] = topo.second;
    }
    dir.push(make_pair(s[a], a));
  }
  esq.push(make_pair(s[n - 2], n - 2));
  for (int a = n - 3; a >= 0; a--) {
    topo = esq.top();
    while (!esq.empty() && s[a] >= topo.first) {
      ant[topo.second] = a;
      esq.pop();
      if (!esq.empty()) topo = esq.top();
    }
    if (!esq.empty()) {
      topo = esq.top();
      prox[a] = topo.second;
    }
    esq.push(make_pair(s[a], a));
  }
  for (int a = n - 2; a >= 0; a--) {
    if (prox[a] != n - 1 && s[prox[a]] == s[a]) {
      prox[a] = prox[prox[a]];
    }
  }
  for (int a = 0; a < m; a++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    long long res = 0;
    for (int b = l; b <= r - 1; b++) {
      res += (b - max(ant[b], l - 1)) * (min(prox[b], r) - b) * s[b];
    }
    cout << res;
    printf("\n");
  }
  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>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
const int MAXN = 500010;
long long s[MAXN], n, m, prox[MAXN], ant[MAXN];
stack<pair<long long, long long> > dir, esq;
int main() {
  cin >> n >> m;
  memset(ant, -1, sizeof(ant));
  for (int a = 0; a <= n; a++) prox[a] = n - 1;
  for (int a = 0; a < n; a++) cin >> s[a];
  for (int a = 0; a < n - 1; a++) {
    s[a] = abs(s[a + 1] - s[a]);
  }
  dir.push(make_pair(s[0], 0));
  pair<long long, long long> topo;
  for (int a = 1; a < n - 1; a++) {
    topo = dir.top();
    while (!dir.empty() && s[a] >= topo.first) {
      prox[topo.second] = a;
      dir.pop();
      if (!dir.empty()) topo = dir.top();
    }
    if (!dir.empty()) {
      topo = dir.top();
      ant[a] = topo.second;
    }
    dir.push(make_pair(s[a], a));
  }
  esq.push(make_pair(s[n - 2], n - 2));
  for (int a = n - 3; a >= 0; a--) {
    topo = esq.top();
    while (!esq.empty() && s[a] >= topo.first) {
      ant[topo.second] = a;
      esq.pop();
      if (!esq.empty()) topo = esq.top();
    }
    if (!esq.empty()) {
      topo = esq.top();
      prox[a] = topo.second;
    }
    esq.push(make_pair(s[a], a));
  }
  for (int a = n - 2; a >= 0; a--) {
    if (prox[a] != n - 1 && s[prox[a]] == s[a]) {
      prox[a] = prox[prox[a]];
    }
  }
  for (int a = 0; a < m; a++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    long long res = 0;
    for (int b = l; b <= r - 1; b++) {
      res += (b - max(ant[b], l - 1)) * (min(prox[b], r) - b) * s[b];
    }
    cout << res;
    printf("\n");
  }
  return 0;
}

```