#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
struct point {
  long long l;
  long long r;
};
point v[maxn];
long long ch[maxn];
long long a[maxn];
long long premax[maxn];
stack<int> st;
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    ch[i - 1] = abs(a[i] - a[i - 1]);
  }
  for (int i = 1; i < n; i++) {
    while (!st.empty()) {
      int w = st.top();
      if (ch[i] > ch[w]) {
        v[w].r = i;
        st.pop();
      } else
        break;
    }
    st.push(i);
  }
  while (!st.empty()) {
    int w = st.top();
    st.pop();
    v[w].r = n;
  }
  for (int i = n - 1; i > 0; i--) {
    while (!st.empty()) {
      int w = st.top();
      if (ch[i] >= ch[w]) {
        v[w].l = i + 1;
        st.pop();
      } else
        break;
    }
    st.push(i);
  }
  while (!st.empty()) {
    int w = st.top();
    st.pop();
    v[w].l = 1;
  }
  while (q--) {
    long long l, r;
    scanf("%I64d%I64d", &l, &r);
    long long ans = 0;
    for (int i = l; i <= r; i++) {
      ans += ch[i] * ((i - max(l, v[i].l) + 1) * (min(r, v[i].r) - i));
    }
    printf("%I64d\n", ans);
  }
  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 int maxn = 100005;
struct point {
  long long l;
  long long r;
};
point v[maxn];
long long ch[maxn];
long long a[maxn];
long long premax[maxn];
stack<int> st;
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    ch[i - 1] = abs(a[i] - a[i - 1]);
  }
  for (int i = 1; i < n; i++) {
    while (!st.empty()) {
      int w = st.top();
      if (ch[i] > ch[w]) {
        v[w].r = i;
        st.pop();
      } else
        break;
    }
    st.push(i);
  }
  while (!st.empty()) {
    int w = st.top();
    st.pop();
    v[w].r = n;
  }
  for (int i = n - 1; i > 0; i--) {
    while (!st.empty()) {
      int w = st.top();
      if (ch[i] >= ch[w]) {
        v[w].l = i + 1;
        st.pop();
      } else
        break;
    }
    st.push(i);
  }
  while (!st.empty()) {
    int w = st.top();
    st.pop();
    v[w].l = 1;
  }
  while (q--) {
    long long l, r;
    scanf("%I64d%I64d", &l, &r);
    long long ans = 0;
    for (int i = l; i <= r; i++) {
      ans += ch[i] * ((i - max(l, v[i].l) + 1) * (min(r, v[i].r) - i));
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```