#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int INF = (1 << 30) - 1;
const int MAXN = 100010;
int n, q;
int A[MAXN], B[MAXN];
int L[MAXN], R[MAXN];
set<pair<int, int> > st;
set<pair<int, int> >::iterator it;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &A[i]);
  }
  for (int i = 1; i < n; ++i) {
    B[i] = abs(A[i + 1] - A[i]);
  }
  for (int i = 1; i < n; ++i) {
    L[i] = 1;
    R[i] = n - 1;
  }
  for (int i = 1; i < n; ++i) {
    st.insert(make_pair(B[i], i));
    it = st.find(make_pair(B[i], i));
    vector<pair<int, int> > g;
    while (1) {
      if (it == st.begin()) break;
      it--;
      if ((*it).first <= B[i]) {
        R[(*it).second] = i - 1;
        g.push_back((*it));
      }
    }
    for (int j = 0; j < g.size(); ++j) st.erase(g[j]);
  }
  st.clear();
  for (int i = n - 1; i >= 1; --i) {
    st.insert(make_pair(B[i], i));
    it = st.find(make_pair(B[i], i));
    vector<pair<int, int> > g;
    while (1) {
      if (it == st.begin()) break;
      it--;
      if ((*it).first < B[i]) {
        L[(*it).second] = i + 1;
        g.push_back((*it));
      }
    }
    for (int j = 0; j < g.size(); ++j) st.erase(g[j]);
  }
  for (int i = 1; i <= q; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    long long ans = 0;
    for (int j = a; j < b; ++j) {
      int l = max(a, L[j]);
      int r = min(b - 1, R[j]);
      long long num;
      if (l == r)
        num = 1;
      else
        num = 1ll * (r - j + 1) * (j - l + 1);
      ans += 1ll * B[j] * num;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### 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 eps = 1e-8;
const int INF = (1 << 30) - 1;
const int MAXN = 100010;
int n, q;
int A[MAXN], B[MAXN];
int L[MAXN], R[MAXN];
set<pair<int, int> > st;
set<pair<int, int> >::iterator it;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &A[i]);
  }
  for (int i = 1; i < n; ++i) {
    B[i] = abs(A[i + 1] - A[i]);
  }
  for (int i = 1; i < n; ++i) {
    L[i] = 1;
    R[i] = n - 1;
  }
  for (int i = 1; i < n; ++i) {
    st.insert(make_pair(B[i], i));
    it = st.find(make_pair(B[i], i));
    vector<pair<int, int> > g;
    while (1) {
      if (it == st.begin()) break;
      it--;
      if ((*it).first <= B[i]) {
        R[(*it).second] = i - 1;
        g.push_back((*it));
      }
    }
    for (int j = 0; j < g.size(); ++j) st.erase(g[j]);
  }
  st.clear();
  for (int i = n - 1; i >= 1; --i) {
    st.insert(make_pair(B[i], i));
    it = st.find(make_pair(B[i], i));
    vector<pair<int, int> > g;
    while (1) {
      if (it == st.begin()) break;
      it--;
      if ((*it).first < B[i]) {
        L[(*it).second] = i + 1;
        g.push_back((*it));
      }
    }
    for (int j = 0; j < g.size(); ++j) st.erase(g[j]);
  }
  for (int i = 1; i <= q; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    long long ans = 0;
    for (int j = a; j < b; ++j) {
      int l = max(a, L[j]);
      int r = min(b - 1, R[j]);
      long long num;
      if (l == r)
        num = 1;
      else
        num = 1ll * (r - j + 1) * (j - l + 1);
      ans += 1ll * B[j] * num;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```