#include <bits/stdc++.h>
using namespace std;
mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
const int N = 1e6 + 200;
const int M = 1e5;
const int mod = 1e9 + 7;
const int inf = 2e9 + 3;
const long long INF = 1e18;
const long double pi2 = 2.0 * 3.141592;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
void files() {}
int n, q, l, r, a[N];
stack<pair<int, pair<long long, long long>>> st;
int main() {
  files();
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  while (q--) {
    cin >> l >> r;
    while ((int)st.size()) st.pop();
    st.push({inf, {0, l}});
    long long ans = 0;
    for (int i = l + 1, x; i <= r; ++i) {
      x = abs(a[i] - a[i - 1]);
      while (st.top().first <= x) {
        st.pop();
      }
      st.push({x,
               {st.top().second.first + x * 1LL * (i - st.top().second.second),
                i}});
      ans += st.top().second.first;
    }
    cout << ans << '\n';
  }
}

### Prompt

Generate 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;
mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());
const int N = 1e6 + 200;
const int M = 1e5;
const int mod = 1e9 + 7;
const int inf = 2e9 + 3;
const long long INF = 1e18;
const long double pi2 = 2.0 * 3.141592;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
void files() {}
int n, q, l, r, a[N];
stack<pair<int, pair<long long, long long>>> st;
int main() {
  files();
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  while (q--) {
    cin >> l >> r;
    while ((int)st.size()) st.pop();
    st.push({inf, {0, l}});
    long long ans = 0;
    for (int i = l + 1, x; i <= r; ++i) {
      x = abs(a[i] - a[i - 1]);
      while (st.top().first <= x) {
        st.pop();
      }
      st.push({x,
               {st.top().second.first + x * 1LL * (i - st.top().second.second),
                i}});
      ans += st.top().second.first;
    }
    cout << ans << '\n';
  }
}

```