#include <bits/stdc++.h>
using namespace std;
const int size1 = 4e5;
const int INF = 1e9;
const double eps = 1e-5;
const long long mod = 1e9 + 7;
vector<int> v;
static int tree[size1];
void build_tree(int tl, int tr, int pos) {
  if (tl == tr) {
    tree[pos] = v[tl];
  } else {
    int tm = (tl + tr) / 2;
    build_tree(tl, tm, pos * 2);
    build_tree(tm + 1, tr, pos * 2 + 1);
    tree[pos] = max(tree[pos * 2], tree[pos * 2 + 1]);
  }
}
int max1(int tl, int tr, int pos, int l, int r) {
  if (l <= tl && tr <= r) {
    return tree[pos];
  } else {
    int tm = (tl + tr) / 2;
    int el = 0;
    if (max(l, tl) <= min(r, tm)) {
      el = max(el, max1(tl, tm, pos * 2, l, r));
    }
    if (max(l, tm + 1) <= min(r, tr)) {
      el = max(el, max1(tm + 1, tr, pos * 2 + 1, l, r));
    }
    return el;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  static int mas[size1];
  int i;
  for (i = 1; i <= n; i++) {
    cin >> mas[i];
    v.push_back(abs(mas[i] - mas[i - 1]));
  }
  static int pred[size1];
  build_tree(1, n, 1);
  for (i = 1; i <= n; i++) {
    int l = 0, r = i;
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (max1(1, n, 1, m, i - 1) >= v[i]) {
        l = m;
      } else {
        r = m;
      }
    }
    pred[i] = l;
  }
  static long long pref[size1];
  for (i = 1; i <= q; i++) {
    int l, r;
    cin >> l >> r;
    if (l == r) {
      cout << 0 << "\n";
      continue;
    }
    int l1 = l;
    int r1 = r - 1;
    long long ans = 0;
    for (int j = l1; j <= r1; j++) {
      if (pred[j] < l1) {
        pref[j] = ((long long)v[j]) * (j - l1 + 1);
      } else {
        pref[j] = ((long long)v[j]) * (j - pred[j]) + pref[pred[j]];
      }
    }
    for (int j = l1; j <= r1; j++) {
      ans += pref[j];
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 size1 = 4e5;
const int INF = 1e9;
const double eps = 1e-5;
const long long mod = 1e9 + 7;
vector<int> v;
static int tree[size1];
void build_tree(int tl, int tr, int pos) {
  if (tl == tr) {
    tree[pos] = v[tl];
  } else {
    int tm = (tl + tr) / 2;
    build_tree(tl, tm, pos * 2);
    build_tree(tm + 1, tr, pos * 2 + 1);
    tree[pos] = max(tree[pos * 2], tree[pos * 2 + 1]);
  }
}
int max1(int tl, int tr, int pos, int l, int r) {
  if (l <= tl && tr <= r) {
    return tree[pos];
  } else {
    int tm = (tl + tr) / 2;
    int el = 0;
    if (max(l, tl) <= min(r, tm)) {
      el = max(el, max1(tl, tm, pos * 2, l, r));
    }
    if (max(l, tm + 1) <= min(r, tr)) {
      el = max(el, max1(tm + 1, tr, pos * 2 + 1, l, r));
    }
    return el;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  static int mas[size1];
  int i;
  for (i = 1; i <= n; i++) {
    cin >> mas[i];
    v.push_back(abs(mas[i] - mas[i - 1]));
  }
  static int pred[size1];
  build_tree(1, n, 1);
  for (i = 1; i <= n; i++) {
    int l = 0, r = i;
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (max1(1, n, 1, m, i - 1) >= v[i]) {
        l = m;
      } else {
        r = m;
      }
    }
    pred[i] = l;
  }
  static long long pref[size1];
  for (i = 1; i <= q; i++) {
    int l, r;
    cin >> l >> r;
    if (l == r) {
      cout << 0 << "\n";
      continue;
    }
    int l1 = l;
    int r1 = r - 1;
    long long ans = 0;
    for (int j = l1; j <= r1; j++) {
      if (pred[j] < l1) {
        pref[j] = ((long long)v[j]) * (j - l1 + 1);
      } else {
        pref[j] = ((long long)v[j]) * (j - pred[j]) + pref[pred[j]];
      }
    }
    for (int j = l1; j <= r1; j++) {
      ans += pref[j];
    }
    cout << ans << "\n";
  }
  return 0;
}

```