#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 1e18;
const int N = 121234;
int n, q;
int a[N];
int f[N];
int l[N], r[N];
int main() {
  while (scanf("%d%d", &n, &q) != EOF) {
    int i, j;
    for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (i = 1; i <= n - 1; ++i) f[i] = abs(a[i] - a[i + 1]);
    memset(l, 0, sizeof(l));
    memset(r, 0, sizeof(r));
    stack<pair<int, int> > st;
    for (i = 1; i <= n - 1; ++i) {
      while (!st.empty()) {
        pair<int, int> cur = st.top();
        if (f[i] >= cur.second)
          st.pop();
        else
          break;
      }
      if (st.empty())
        l[i] = i - 1;
      else {
        pair<int, int> cur = st.top();
        l[i] = i - cur.first - 1;
      }
      st.push(make_pair(i, f[i]));
    }
    while (!st.empty()) st.pop();
    for (i = n - 1; i >= 1; --i) {
      while (!st.empty()) {
        pair<int, int> cur = st.top();
        if (f[i] > cur.second)
          st.pop();
        else
          break;
      }
      if (st.empty())
        r[i] = n - 1 - i;
      else {
        pair<int, int> cur = st.top();
        r[i] = cur.first - 1 - i;
      }
      st.push(make_pair(i, f[i]));
    }
    while (q--) {
      int s, t;
      scanf("%d%d", &s, &t);
      long long ans = 0;
      t--;
      for (i = s; i <= t; ++i) {
        ans += f[i] * (1 + 1LL * min(i - s, l[i]) * min(t - i, r[i]) +
                       min(i - s, l[i]) + min(t - i, r[i]));
      }
      printf("%I64d\n", ans);
    }
  }
}

### 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 inf = 0x3f3f3f3f;
const long long INF = 1e18;
const int N = 121234;
int n, q;
int a[N];
int f[N];
int l[N], r[N];
int main() {
  while (scanf("%d%d", &n, &q) != EOF) {
    int i, j;
    for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (i = 1; i <= n - 1; ++i) f[i] = abs(a[i] - a[i + 1]);
    memset(l, 0, sizeof(l));
    memset(r, 0, sizeof(r));
    stack<pair<int, int> > st;
    for (i = 1; i <= n - 1; ++i) {
      while (!st.empty()) {
        pair<int, int> cur = st.top();
        if (f[i] >= cur.second)
          st.pop();
        else
          break;
      }
      if (st.empty())
        l[i] = i - 1;
      else {
        pair<int, int> cur = st.top();
        l[i] = i - cur.first - 1;
      }
      st.push(make_pair(i, f[i]));
    }
    while (!st.empty()) st.pop();
    for (i = n - 1; i >= 1; --i) {
      while (!st.empty()) {
        pair<int, int> cur = st.top();
        if (f[i] > cur.second)
          st.pop();
        else
          break;
      }
      if (st.empty())
        r[i] = n - 1 - i;
      else {
        pair<int, int> cur = st.top();
        r[i] = cur.first - 1 - i;
      }
      st.push(make_pair(i, f[i]));
    }
    while (q--) {
      int s, t;
      scanf("%d%d", &s, &t);
      long long ans = 0;
      t--;
      for (i = s; i <= t; ++i) {
        ans += f[i] * (1 + 1LL * min(i - s, l[i]) * min(t - i, r[i]) +
                       min(i - s, l[i]) + min(t - i, r[i]));
      }
      printf("%I64d\n", ans);
    }
  }
}

```