#include <bits/stdc++.h>
using namespace std;
const long long S = (1 << 17);
long long a, q, x, y, id[S];
pair<long long, long long> joint[S];
vector<long long> X;
vector<pair<long long, long long> > Y;
long long foo(pair<long long, long long> H) {
  if (H.second < H.first) return (long long)0;
  long long Z = abs(H.first - H.second) + 1;
  return Z * (Z + 1) / 2;
}
long long Ans(long long poc, long long kon, long long ind) {
  long long my_ans = (long long)0;
  for (auto e : Y) {
    long long val = e.first;
    long long pos = e.second;
    if (pos < poc or pos > kon) continue;
    pair<long long, long long> range = {max(joint[pos].first, poc),
                                        min(kon, joint[pos].second)};
    long long to_add = 0;
    my_ans += (foo(range) - foo({range.first, pos - 1}) -
               foo({pos + 1, range.second})) *
              val;
  }
  return my_ans;
}
int main() {
  scanf("%lld%lld%lld", &a, &q, &x);
  for (int i = 0; i < a - 1; i++) {
    scanf("%lld", &y);
    X.push_back(abs(y - x));
    Y.push_back({abs(y - x), i});
    x = y;
  }
  stack<pair<long long, long long> > Q;
  sort(Y.begin(), Y.end());
  for (int i = 0; i < Y.size(); i++) id[Y[i].second] = i;
  for (int i = 0; i < X.size(); i++) {
    while (!Q.empty() and Q.top().second <= X[i]) {
      joint[Q.top().first] = {0, i - 1};
      Q.pop();
    }
    Q.push({i, X[i]});
  }
  while (!Q.empty()) {
    long long pos = Q.top().first;
    Q.pop();
    joint[pos] = {0, X.size() - 1};
  }
  for (int i = X.size() - 1; i >= 0; i--) {
    while (!Q.empty() and Q.top().second < X[i]) {
      joint[Q.top().first] = {i + 1, joint[Q.top().first].second};
      Q.pop();
    }
    Q.push({i, X[i]});
  }
  while (!Q.empty()) {
    long long pos = Q.top().first;
    Q.pop();
    joint[pos] = {0, joint[pos].second};
  }
  for (int i = 0; i < q; i++) {
    scanf("%lld%lld", &x, &y);
    printf("%lld\n", Ans(x - 1, y - 2, (long long)i));
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
const long long S = (1 << 17);
long long a, q, x, y, id[S];
pair<long long, long long> joint[S];
vector<long long> X;
vector<pair<long long, long long> > Y;
long long foo(pair<long long, long long> H) {
  if (H.second < H.first) return (long long)0;
  long long Z = abs(H.first - H.second) + 1;
  return Z * (Z + 1) / 2;
}
long long Ans(long long poc, long long kon, long long ind) {
  long long my_ans = (long long)0;
  for (auto e : Y) {
    long long val = e.first;
    long long pos = e.second;
    if (pos < poc or pos > kon) continue;
    pair<long long, long long> range = {max(joint[pos].first, poc),
                                        min(kon, joint[pos].second)};
    long long to_add = 0;
    my_ans += (foo(range) - foo({range.first, pos - 1}) -
               foo({pos + 1, range.second})) *
              val;
  }
  return my_ans;
}
int main() {
  scanf("%lld%lld%lld", &a, &q, &x);
  for (int i = 0; i < a - 1; i++) {
    scanf("%lld", &y);
    X.push_back(abs(y - x));
    Y.push_back({abs(y - x), i});
    x = y;
  }
  stack<pair<long long, long long> > Q;
  sort(Y.begin(), Y.end());
  for (int i = 0; i < Y.size(); i++) id[Y[i].second] = i;
  for (int i = 0; i < X.size(); i++) {
    while (!Q.empty() and Q.top().second <= X[i]) {
      joint[Q.top().first] = {0, i - 1};
      Q.pop();
    }
    Q.push({i, X[i]});
  }
  while (!Q.empty()) {
    long long pos = Q.top().first;
    Q.pop();
    joint[pos] = {0, X.size() - 1};
  }
  for (int i = X.size() - 1; i >= 0; i--) {
    while (!Q.empty() and Q.top().second < X[i]) {
      joint[Q.top().first] = {i + 1, joint[Q.top().first].second};
      Q.pop();
    }
    Q.push({i, X[i]});
  }
  while (!Q.empty()) {
    long long pos = Q.top().first;
    Q.pop();
    joint[pos] = {0, joint[pos].second};
  }
  for (int i = 0; i < q; i++) {
    scanf("%lld%lld", &x, &y);
    printf("%lld\n", Ans(x - 1, y - 2, (long long)i));
  }
  return 0;
}

```