#include <bits/stdc++.h>
using namespace std;
const int N = int(1e5) + 10;
const int LOGN = 20;
int A[N], B[N], msb[N];
int DP[LOGN][N];
void build_msb() {
  int mx = -1;
  for (int i = 0; i < N; i++) {
    if (i >= (1 << (mx + 1))) mx++;
    msb[i] = mx;
  }
}
int MAX(int l, int r) {
  int idx = msb[r - l + 1];
  int sub = (1 << idx);
  int ret = ((A[DP[idx][l]] > A[DP[idx][r - sub + 1]]) ? DP[idx][l]
                                                       : DP[idx][r - sub + 1]);
  return ret;
}
long long int solve(int l, int r) {
  if (l >= r) return 0;
  int mx = MAX(l, r - 1);
  ;
  long long int count = (mx - l + 1) * 1ll * (r - mx);
  long long int add = A[mx] * count;
  ;
  return add + solve(l, mx) + solve(mx + 1, r);
}
int main() {
  build_msb();
  int n;
  scanf("%d", &n);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) scanf("%d", B + i);
  for (int i = 1; i < n; i++) A[i] = abs(B[i + 1] - B[i]);
  for (int i = 1; i <= n; i++) DP[0][i] = i;
  for (int i = 1; i < LOGN; i++)
    for (int j = 1; j <= n; j++) {
      int l = j, r = j + (1 << (i - 1));
      if (r > n)
        DP[i][j] = DP[i - 1][j];
      else
        DP[i][j] =
            (A[DP[i - 1][l]] > A[DP[i - 1][r]] ? DP[i - 1][l] : DP[i - 1][r]);
    }
  while (q--) {
    int l, r;
    scanf("%d", &l);
    scanf("%d", &r);
    printf("%lld\n", solve(l, r));
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = int(1e5) + 10;
const int LOGN = 20;
int A[N], B[N], msb[N];
int DP[LOGN][N];
void build_msb() {
  int mx = -1;
  for (int i = 0; i < N; i++) {
    if (i >= (1 << (mx + 1))) mx++;
    msb[i] = mx;
  }
}
int MAX(int l, int r) {
  int idx = msb[r - l + 1];
  int sub = (1 << idx);
  int ret = ((A[DP[idx][l]] > A[DP[idx][r - sub + 1]]) ? DP[idx][l]
                                                       : DP[idx][r - sub + 1]);
  return ret;
}
long long int solve(int l, int r) {
  if (l >= r) return 0;
  int mx = MAX(l, r - 1);
  ;
  long long int count = (mx - l + 1) * 1ll * (r - mx);
  long long int add = A[mx] * count;
  ;
  return add + solve(l, mx) + solve(mx + 1, r);
}
int main() {
  build_msb();
  int n;
  scanf("%d", &n);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) scanf("%d", B + i);
  for (int i = 1; i < n; i++) A[i] = abs(B[i + 1] - B[i]);
  for (int i = 1; i <= n; i++) DP[0][i] = i;
  for (int i = 1; i < LOGN; i++)
    for (int j = 1; j <= n; j++) {
      int l = j, r = j + (1 << (i - 1));
      if (r > n)
        DP[i][j] = DP[i - 1][j];
      else
        DP[i][j] =
            (A[DP[i - 1][l]] > A[DP[i - 1][r]] ? DP[i - 1][l] : DP[i - 1][r]);
    }
  while (q--) {
    int l, r;
    scanf("%d", &l);
    scanf("%d", &r);
    printf("%lld\n", solve(l, r));
  }
  return 0;
}

```