#include <bits/stdc++.h>
using namespace std;
const double PI = acos(0) * 2;
const double EPS = 1e-8;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
const int oo = 1e9;
const double foo = 1e30;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcounll(s);
}
int a[100100], n;
long long solve(int l, int r) {
  vector<long long> d;
  int L[100100], R[100100];
  stack<pair<int, int> > st;
  memset(L, 0, sizeof L);
  memset(R, 0, sizeof R);
  d.clear();
  d.push_back(oo);
  for (int i = l + 1; i <= r; i++) {
    int x = abs(a[i] - a[i - 1]);
    d.push_back(x);
  }
  d.push_back(oo);
  st.push(make_pair(d[0], 0));
  int N = ((int)(d).size());
  for (int i = 1; i < ((int)(d).size()); i++) {
    while (st.top().first < d[i]) st.pop();
    L[i] = st.top().second;
    st.push(make_pair(d[i], i));
  }
  while (!st.empty()) st.pop();
  st.push(make_pair(d[N - 1], N - 1));
  for (int i = N - 2; i >= 1; i--) {
    while (!st.empty() && st.top().first <= d[i]) st.pop();
    R[i] = st.top().second;
    st.push(make_pair(d[i], i));
  }
  long long ret = 0;
  for (int i = 1; i <= N - 2; i++) {
    long long l = L[i] + 1;
    long long r = R[i] - 1;
    l = i - l + 1;
    r = r - i + 1;
    long long x = (l * r) * d[i];
    ret += x;
  }
  return ret;
}
int main() {
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    printf("%I64d\n", solve(l, r));
  }
}

### 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 double PI = acos(0) * 2;
const double EPS = 1e-8;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
const int oo = 1e9;
const double foo = 1e30;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcounll(s);
}
int a[100100], n;
long long solve(int l, int r) {
  vector<long long> d;
  int L[100100], R[100100];
  stack<pair<int, int> > st;
  memset(L, 0, sizeof L);
  memset(R, 0, sizeof R);
  d.clear();
  d.push_back(oo);
  for (int i = l + 1; i <= r; i++) {
    int x = abs(a[i] - a[i - 1]);
    d.push_back(x);
  }
  d.push_back(oo);
  st.push(make_pair(d[0], 0));
  int N = ((int)(d).size());
  for (int i = 1; i < ((int)(d).size()); i++) {
    while (st.top().first < d[i]) st.pop();
    L[i] = st.top().second;
    st.push(make_pair(d[i], i));
  }
  while (!st.empty()) st.pop();
  st.push(make_pair(d[N - 1], N - 1));
  for (int i = N - 2; i >= 1; i--) {
    while (!st.empty() && st.top().first <= d[i]) st.pop();
    R[i] = st.top().second;
    st.push(make_pair(d[i], i));
  }
  long long ret = 0;
  for (int i = 1; i <= N - 2; i++) {
    long long l = L[i] + 1;
    long long r = R[i] - 1;
    l = i - l + 1;
    r = r - i + 1;
    long long x = (l * r) * d[i];
    ret += x;
  }
  return ret;
}
int main() {
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    printf("%I64d\n", solve(l, r));
  }
}

```