#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
const int MOD = 1000000007;
const int N = 1E5 + 15;
const int maxn = 2e5 + 14;
const int INF = (1 << 30) - 1;
const int letter = 130;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, q;
long long a[N], b[N];
long long L[N], R[N];
int s[N], top;
void solve(int l, int r) {
  int cnt = 0;
  for (int i = l; i < r; i++) b[++cnt] = a[i];
  b[0] = INF;
  b[cnt + 1] = INF;
  s[top = 0] = 0;
  for (int i = 1; i <= cnt + 1; i++) {
    if (b[i] <= b[s[top]])
      s[++top] = i;
    else {
      while (top >= 0 && b[s[top]] < b[i]) R[s[top]] = i - s[top], top--;
      s[++top] = i;
    }
  }
  s[top = 0] = cnt + 1;
  for (int i = cnt; i >= 0; i--) {
    if (b[i] < b[s[top]])
      s[++top] = i;
    else {
      while (top >= 0 && b[s[top]] <= b[i]) L[s[top]] = s[top] - i, top--;
      s[++top] = i;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= cnt; i++) {
    ans += L[i] * R[i] * b[i];
  }
  printf("%I64d\n", ans);
}
int main() {
  int x, y;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%I64d", a + i);
  for (int i = 1; i < n; i++) a[i] = abs(a[i + 1] - a[i]);
  while (q--) {
    scanf("%d%d", &x, &y);
    solve(x, y);
  }
  return 0;
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:102400000,102400000")
const int MOD = 1000000007;
const int N = 1E5 + 15;
const int maxn = 2e5 + 14;
const int INF = (1 << 30) - 1;
const int letter = 130;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, q;
long long a[N], b[N];
long long L[N], R[N];
int s[N], top;
void solve(int l, int r) {
  int cnt = 0;
  for (int i = l; i < r; i++) b[++cnt] = a[i];
  b[0] = INF;
  b[cnt + 1] = INF;
  s[top = 0] = 0;
  for (int i = 1; i <= cnt + 1; i++) {
    if (b[i] <= b[s[top]])
      s[++top] = i;
    else {
      while (top >= 0 && b[s[top]] < b[i]) R[s[top]] = i - s[top], top--;
      s[++top] = i;
    }
  }
  s[top = 0] = cnt + 1;
  for (int i = cnt; i >= 0; i--) {
    if (b[i] < b[s[top]])
      s[++top] = i;
    else {
      while (top >= 0 && b[s[top]] <= b[i]) L[s[top]] = s[top] - i, top--;
      s[++top] = i;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= cnt; i++) {
    ans += L[i] * R[i] * b[i];
  }
  printf("%I64d\n", ans);
}
int main() {
  int x, y;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%I64d", a + i);
  for (int i = 1; i < n; i++) a[i] = abs(a[i + 1] - a[i]);
  while (q--) {
    scanf("%d%d", &x, &y);
    solve(x, y);
  }
  return 0;
}

```