#include <bits/stdc++.h>
using namespace std;
const int N = 110000;
int i, j, k, l, m, n, o, p, mid, f[20][N], b[N], is[N], a[N], L[N], R[N], lg[N];
struct arr {
  int x, y;
  arr() {}
  arr(int X, int Y) {
    x = X;
    y = Y;
  }
  inline bool operator<(const arr &t) const {
    return (x < t.x) || ((x == t.x) && (y < t.y));
  }
};
map<arr, int> F;
inline int read() {
  int o = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while ((c >= '0') && (c <= '9')) {
    o = (o << 1) + (o << 3) + c - '0';
    c = getchar();
  }
  return o;
}
inline void Init() {
  n = read();
  m = read();
  for (i = 1; i <= n; i++) a[i] = read();
  for (i = 1; i <= n - 1; i++) {
    b[i] = a[i + 1] - a[i];
    if (b[i] < 0) b[i] = -b[i];
  }
  lg[1] = 0;
  for (i = 2; i <= n; i++) lg[i] = lg[i - 1] + ((1 << lg[i - 1] + 1) == i);
}
inline int get(int L, int R) {
  int len = R - L + 1, k;
  k = lg[len];
  return max(f[k][L], f[k][R - (1 << k) + 1]);
}
inline void Prepare() {
  for (i = 1; i <= n - 1; i++) f[0][i] = b[i];
  for (i = 1; (1 << i) < n; i++)
    for (int j = 1; j < n; j++) {
      o = j + (1 << i - 1);
      if (o >= n) o = n - 1;
      f[i][j] = max(f[i - 1][j], f[i - 1][o]);
    }
  for (i = 1; i <= n - 1; i++) {
    o = 1;
    p = i - 1;
    while (o <= p) {
      mid = o + p >> 1;
      if (get(mid, i - 1) < b[i])
        p = mid - 1;
      else
        o = mid + 1;
    }
    L[i] = o;
    o = i + 1;
    p = n - 1;
    while (o <= p) {
      mid = o + p >> 1;
      if (get(i + 1, mid) <= b[i])
        o = mid + 1;
      else
        p = mid - 1;
    }
    R[i] = p + 1;
  }
}
inline void Solve() {
  long long l, r;
  long long ans = 0, t1, t2;
  while (m--) {
    o = read();
    p = read();
    ans = 0;
    for (i = 1; i <= n - 1; i++)
      if ((o <= i) && (i + 1 <= p)) {
        l = max(L[i], o);
        r = min(R[i], p);
        ans += (long long)(i - l + 1) * (long long)(r - i) * (long long)(b[i]);
      }
    printf("%I64d\n", ans);
  }
}
int main() {
  Init();
  Prepare();
  Solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 110000;
int i, j, k, l, m, n, o, p, mid, f[20][N], b[N], is[N], a[N], L[N], R[N], lg[N];
struct arr {
  int x, y;
  arr() {}
  arr(int X, int Y) {
    x = X;
    y = Y;
  }
  inline bool operator<(const arr &t) const {
    return (x < t.x) || ((x == t.x) && (y < t.y));
  }
};
map<arr, int> F;
inline int read() {
  int o = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while ((c >= '0') && (c <= '9')) {
    o = (o << 1) + (o << 3) + c - '0';
    c = getchar();
  }
  return o;
}
inline void Init() {
  n = read();
  m = read();
  for (i = 1; i <= n; i++) a[i] = read();
  for (i = 1; i <= n - 1; i++) {
    b[i] = a[i + 1] - a[i];
    if (b[i] < 0) b[i] = -b[i];
  }
  lg[1] = 0;
  for (i = 2; i <= n; i++) lg[i] = lg[i - 1] + ((1 << lg[i - 1] + 1) == i);
}
inline int get(int L, int R) {
  int len = R - L + 1, k;
  k = lg[len];
  return max(f[k][L], f[k][R - (1 << k) + 1]);
}
inline void Prepare() {
  for (i = 1; i <= n - 1; i++) f[0][i] = b[i];
  for (i = 1; (1 << i) < n; i++)
    for (int j = 1; j < n; j++) {
      o = j + (1 << i - 1);
      if (o >= n) o = n - 1;
      f[i][j] = max(f[i - 1][j], f[i - 1][o]);
    }
  for (i = 1; i <= n - 1; i++) {
    o = 1;
    p = i - 1;
    while (o <= p) {
      mid = o + p >> 1;
      if (get(mid, i - 1) < b[i])
        p = mid - 1;
      else
        o = mid + 1;
    }
    L[i] = o;
    o = i + 1;
    p = n - 1;
    while (o <= p) {
      mid = o + p >> 1;
      if (get(i + 1, mid) <= b[i])
        o = mid + 1;
      else
        p = mid - 1;
    }
    R[i] = p + 1;
  }
}
inline void Solve() {
  long long l, r;
  long long ans = 0, t1, t2;
  while (m--) {
    o = read();
    p = read();
    ans = 0;
    for (i = 1; i <= n - 1; i++)
      if ((o <= i) && (i + 1 <= p)) {
        l = max(L[i], o);
        r = min(R[i], p);
        ans += (long long)(i - l + 1) * (long long)(r - i) * (long long)(b[i]);
      }
    printf("%I64d\n", ans);
  }
}
int main() {
  Init();
  Prepare();
  Solve();
  return 0;
}

```