#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5, maxlgn = 17;
int n, a[maxn];
int delta[maxn];
int st[maxn][maxlgn];
int h[maxn];
int Max(int i, int j) { return delta[i] < delta[j] ? j : i; }
void InitSt(int n) {
  for (int i = 0; i < n; ++i) {
    st[i][0] = i;
  }
  for (int k = 1; 1 << k <= n; ++k) {
    for (int i = 0; i + (1 << k) <= n; ++i) {
      st[i][k] = Max(st[i][k - 1], st[i + (1 << k - 1)][k - 1]);
    }
  }
  h[0] = 0;
  for (int i = 1; i <= n; ++i) {
    h[i] = h[i - 1];
    while (1 << h[i] + 1 <= i) {
      ++h[i];
    }
  }
}
int Query(int l, int r) {
  int k = h[r - l];
  return Max(st[l][k], st[r - (1 << k)][k]);
}
long long Solve(int l, int r) {
  if (r - l <= 0) return 0;
  int mid = Query(l, r);
  long long ans = (long long)(mid - l + 1) * (r - mid) * delta[mid];
  ans += Solve(l, mid);
  ans += Solve(mid + 1, r);
  return ans;
}
int main(void) {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < n - 1; ++i) {
    delta[i] = abs(a[i + 1] - a[i]);
  }
  InitSt(n - 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l;
    long long ans = Solve(l, r - 1);
    printf("%lld\n", ans);
  }
  return 0;
}

### 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;
const int maxn = 1e5, maxlgn = 17;
int n, a[maxn];
int delta[maxn];
int st[maxn][maxlgn];
int h[maxn];
int Max(int i, int j) { return delta[i] < delta[j] ? j : i; }
void InitSt(int n) {
  for (int i = 0; i < n; ++i) {
    st[i][0] = i;
  }
  for (int k = 1; 1 << k <= n; ++k) {
    for (int i = 0; i + (1 << k) <= n; ++i) {
      st[i][k] = Max(st[i][k - 1], st[i + (1 << k - 1)][k - 1]);
    }
  }
  h[0] = 0;
  for (int i = 1; i <= n; ++i) {
    h[i] = h[i - 1];
    while (1 << h[i] + 1 <= i) {
      ++h[i];
    }
  }
}
int Query(int l, int r) {
  int k = h[r - l];
  return Max(st[l][k], st[r - (1 << k)][k]);
}
long long Solve(int l, int r) {
  if (r - l <= 0) return 0;
  int mid = Query(l, r);
  long long ans = (long long)(mid - l + 1) * (r - mid) * delta[mid];
  ans += Solve(l, mid);
  ans += Solve(mid + 1, r);
  return ans;
}
int main(void) {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < n - 1; ++i) {
    delta[i] = abs(a[i + 1] - a[i]);
  }
  InitSt(n - 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l;
    long long ans = Solve(l, r - 1);
    printf("%lld\n", ans);
  }
  return 0;
}

```