#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-11;
int n, q, l, r;
int a[100100];
int b[100100];
signed long long ans;
pair<int, int> t[400400];
int ds;
void build(int ind, int L, int R) {
  if (L == R) {
    t[ind].first = b[L];
    t[ind].second = L;
    if (L == 0) ds = ind;
  } else {
    int M = (L + R) / 2;
    build(ind * 2, L, M);
    build(ind * 2 + 1, M + 1, R);
    if (t[ind * 2].first > t[ind * 2 + 1].first)
      t[ind] = t[ind * 2];
    else
      t[ind] = t[ind * 2 + 1];
  }
}
pair<int, int> mmax(int ind, int l, int r, int L, int R) {
  if (L > R) return make_pair(-1, -1);
  if (L == l && R == r) return t[ind];
  int mm = (l + r) >> 1;
  return max(mmax(ind << 1, l, mm, L, min(R, mm)),
             mmax((ind << 1) + 1, mm + 1, r, max(L, mm + 1), R));
}
void tadd(int first, pair<int, int> zn) {
  first += ds;
  t[first] = zn;
  first /= 2;
  while (first) {
    t[first] = max(zn, t[first]);
    first /= 2;
  }
}
pair<int, int> tsum(int l, int r) {
  pair<int, int> s = make_pair(-1, -1);
  l += ds;
  r += ds;
  while (l <= r) {
    if (l & 1) s = max(s, t[l]);
    if (!(r & 1)) s = max(s, t[r]);
    l = (l + 1) / 2;
    r = (r - 1) / 2;
  }
  return s;
}
int uk, len;
pair<int, int> st[200100], pos;
void proc() {
  l = st[uk].first;
  r = st[uk].second;
  uk++;
  if (l == r) {
    ans += b[l];
    return;
  }
  pos = tsum(l, r);
  if (pos == make_pair(-1, -1)) return;
  ans += 1ll * b[pos.second] * (pos.second - l + 1) * (r - pos.second + 1);
  if (l <= pos.second - 1) st[len++] = make_pair(l, pos.second - 1);
  if (pos.second + 1 <= r) st[len++] = make_pair(pos.second + 1, r);
}
int main() {
  cout << setiosflags(ios::fixed) << setprecision(10);
  cin >> n >> q;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (i > 0) b[i - 1] = abs(a[i] - a[i - 1]);
  }
  ds = 131072;
  for (int i = 0; i < (n - 1); i++) tadd(i, make_pair(b[i], i));
  for (int i = 0; i < (q); i++) {
    cin >> l >> r;
    l--, r--;
    r--;
    ans = 0;
    uk = 0, len = 0;
    st[len++] = make_pair(l, r);
    while (uk < len) proc();
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate 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;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-11;
int n, q, l, r;
int a[100100];
int b[100100];
signed long long ans;
pair<int, int> t[400400];
int ds;
void build(int ind, int L, int R) {
  if (L == R) {
    t[ind].first = b[L];
    t[ind].second = L;
    if (L == 0) ds = ind;
  } else {
    int M = (L + R) / 2;
    build(ind * 2, L, M);
    build(ind * 2 + 1, M + 1, R);
    if (t[ind * 2].first > t[ind * 2 + 1].first)
      t[ind] = t[ind * 2];
    else
      t[ind] = t[ind * 2 + 1];
  }
}
pair<int, int> mmax(int ind, int l, int r, int L, int R) {
  if (L > R) return make_pair(-1, -1);
  if (L == l && R == r) return t[ind];
  int mm = (l + r) >> 1;
  return max(mmax(ind << 1, l, mm, L, min(R, mm)),
             mmax((ind << 1) + 1, mm + 1, r, max(L, mm + 1), R));
}
void tadd(int first, pair<int, int> zn) {
  first += ds;
  t[first] = zn;
  first /= 2;
  while (first) {
    t[first] = max(zn, t[first]);
    first /= 2;
  }
}
pair<int, int> tsum(int l, int r) {
  pair<int, int> s = make_pair(-1, -1);
  l += ds;
  r += ds;
  while (l <= r) {
    if (l & 1) s = max(s, t[l]);
    if (!(r & 1)) s = max(s, t[r]);
    l = (l + 1) / 2;
    r = (r - 1) / 2;
  }
  return s;
}
int uk, len;
pair<int, int> st[200100], pos;
void proc() {
  l = st[uk].first;
  r = st[uk].second;
  uk++;
  if (l == r) {
    ans += b[l];
    return;
  }
  pos = tsum(l, r);
  if (pos == make_pair(-1, -1)) return;
  ans += 1ll * b[pos.second] * (pos.second - l + 1) * (r - pos.second + 1);
  if (l <= pos.second - 1) st[len++] = make_pair(l, pos.second - 1);
  if (pos.second + 1 <= r) st[len++] = make_pair(pos.second + 1, r);
}
int main() {
  cout << setiosflags(ios::fixed) << setprecision(10);
  cin >> n >> q;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (i > 0) b[i - 1] = abs(a[i] - a[i - 1]);
  }
  ds = 131072;
  for (int i = 0; i < (n - 1); i++) tadd(i, make_pair(b[i], i));
  for (int i = 0; i < (q); i++) {
    cin >> l >> r;
    l--, r--;
    r--;
    ans = 0;
    uk = 0, len = 0;
    st[len++] = make_pair(l, r);
    while (uk < len) proc();
    cout << ans << endl;
  }
  return 0;
}

```