#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int n, q;
int in[N];
int diff[N];
vector<pair<int, int> > V;
long long int solve(int from, int to) {
  long long int ret = 0LL;
  for (int i = from; i <= to; ++i) {
    int cur = i;
    while (V.size()) {
      if (diff[V.back().first] > diff[i]) break;
      ret += 1LL * diff[V.back().first] * (i - V.back().first) *
             (V.back().first - V.back().second + 1);
      cur = V.back().second;
      V.pop_back();
    }
    V.push_back({i, cur});
  }
  ++to;
  while (V.size()) {
    ret += 1LL * diff[V.back().first] * (to - V.back().first) *
           (V.back().first - V.back().second + 1);
    V.pop_back();
  }
  return ret;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &in[i]);
  for (int i = 1; i < n; ++i) diff[i] = abs(in[i + 1] - in[i]);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%lld\n", solve(l, r - 1));
  }
  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;
const int N = 1e5 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int n, q;
int in[N];
int diff[N];
vector<pair<int, int> > V;
long long int solve(int from, int to) {
  long long int ret = 0LL;
  for (int i = from; i <= to; ++i) {
    int cur = i;
    while (V.size()) {
      if (diff[V.back().first] > diff[i]) break;
      ret += 1LL * diff[V.back().first] * (i - V.back().first) *
             (V.back().first - V.back().second + 1);
      cur = V.back().second;
      V.pop_back();
    }
    V.push_back({i, cur});
  }
  ++to;
  while (V.size()) {
    ret += 1LL * diff[V.back().first] * (to - V.back().first) *
           (V.back().first - V.back().second + 1);
    V.pop_back();
  }
  return ret;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &in[i]);
  for (int i = 1; i < n; ++i) diff[i] = abs(in[i + 1] - in[i]);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%lld\n", solve(l, r - 1));
  }
  return 0;
}

```