#include <bits/stdc++.h>
using namespace std;
template <class T>
T Bitcnt(T a) {
  int sum = 0;
  while (a) {
    if (a & 1) sum++;
    a /= 2;
  }
  return sum;
}
template <class T>
T Max3(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T Lcm(T a, T b) {
  T tmp = __gcd(a, b);
  return (a / tmp) * b;
}
template <class T>
T Pow(T a, T b) {
  T ans = 1;
  T base = a;
  while (b) {
    if (b & 1) ans = (ans * base);
    base = (base * base);
    b /= 2;
  }
  return ans;
}
long long Bigmod(long long a, long long b) {
  long long res = 1;
  long long pw = a % 1000000007LL;
  while (b > 0) {
    if (b & 1) res = (res * pw) % 1000000007LL;
    pw = (pw * pw) % 1000000007LL;
    b /= 2;
  }
  return res;
}
int a_x[] = {1, -1, 0, 0};
int a_y[] = {0, 0, 1, -1};
long long X, Y;
void extend_euclid(long long a, long long b) {
  if (b == 0) {
    X = a;
    Y = 0;
    return;
  }
  extend_euclid(b, a % b);
  long long x, y;
  x = Y;
  y = X - (a / b) * Y;
  X = x;
  Y = y;
}
long long inverse_modulo(long long a, long long b) {
  extend_euclid(a, b);
  return (X + 1000000007LL) % 1000000007LL;
}
int a[4000000];
pair<long long, int> order[4000000];
long long solve(int l, int r) {
  int cnt = 0;
  if (l == r) return 0;
  order[cnt] = make_pair(1LL << 60, l - 1);
  cnt++;
  long long dp = 0, res = 0;
  for (int i = l + 1; i <= r; i++) {
    long long df = abs(a[i] - a[i - 1]);
    while (df >= order[cnt - 1].first) {
      dp -= (1LL * (order[cnt - 1].second - order[cnt - 2].second) *
             order[cnt - 1].first);
      cnt--;
    }
    order[cnt++] = make_pair(df, i - 1);
    dp += (1LL * (order[cnt - 1].second - order[cnt - 2].second) *
           order[cnt - 1].first);
    res += dp;
  }
  return res;
}
int main() {
  int n, q;
  scanf("%d", &n);
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (q--) {
    int a, b;
    scanf("%d %d", &a, &b);
    long long res = solve(a, b);
    printf("%lld\n", res);
  }
  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>
using namespace std;
template <class T>
T Bitcnt(T a) {
  int sum = 0;
  while (a) {
    if (a & 1) sum++;
    a /= 2;
  }
  return sum;
}
template <class T>
T Max3(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T Lcm(T a, T b) {
  T tmp = __gcd(a, b);
  return (a / tmp) * b;
}
template <class T>
T Pow(T a, T b) {
  T ans = 1;
  T base = a;
  while (b) {
    if (b & 1) ans = (ans * base);
    base = (base * base);
    b /= 2;
  }
  return ans;
}
long long Bigmod(long long a, long long b) {
  long long res = 1;
  long long pw = a % 1000000007LL;
  while (b > 0) {
    if (b & 1) res = (res * pw) % 1000000007LL;
    pw = (pw * pw) % 1000000007LL;
    b /= 2;
  }
  return res;
}
int a_x[] = {1, -1, 0, 0};
int a_y[] = {0, 0, 1, -1};
long long X, Y;
void extend_euclid(long long a, long long b) {
  if (b == 0) {
    X = a;
    Y = 0;
    return;
  }
  extend_euclid(b, a % b);
  long long x, y;
  x = Y;
  y = X - (a / b) * Y;
  X = x;
  Y = y;
}
long long inverse_modulo(long long a, long long b) {
  extend_euclid(a, b);
  return (X + 1000000007LL) % 1000000007LL;
}
int a[4000000];
pair<long long, int> order[4000000];
long long solve(int l, int r) {
  int cnt = 0;
  if (l == r) return 0;
  order[cnt] = make_pair(1LL << 60, l - 1);
  cnt++;
  long long dp = 0, res = 0;
  for (int i = l + 1; i <= r; i++) {
    long long df = abs(a[i] - a[i - 1]);
    while (df >= order[cnt - 1].first) {
      dp -= (1LL * (order[cnt - 1].second - order[cnt - 2].second) *
             order[cnt - 1].first);
      cnt--;
    }
    order[cnt++] = make_pair(df, i - 1);
    dp += (1LL * (order[cnt - 1].second - order[cnt - 2].second) *
           order[cnt - 1].first);
    res += dp;
  }
  return res;
}
int main() {
  int n, q;
  scanf("%d", &n);
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (q--) {
    int a, b;
    scanf("%d %d", &a, &b);
    long long res = solve(a, b);
    printf("%lld\n", res);
  }
  return 0;
}

```