#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int C = 26;
void add(int &x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 1e5 + 10;
long long n, q, a[MAXN], b[MAXN];
int blue[MAXN], red[MAXN], vec[MAXN];
void go(int l, int r) {
  fill(blue + l, blue + r, l - 1);
  fill(red + l, red + r, r);
  int tt = 0;
  for (int i = l; i < r; i++) {
    while (tt && b[vec[tt - 1]] <= b[i]) --tt;
    if (tt) blue[i] = vec[tt - 1];
    vec[tt++] = i;
  }
  tt = 0;
  for (int i = r - 1; i >= l; --i) {
    while (tt && b[vec[tt - 1]] < b[i]) --tt;
    if (tt) red[i] = vec[tt - 1];
    vec[tt++] = i;
  }
  long long ans = 0;
  for (int i = l; i < r; i++) ans += b[i] * (red[i] - i) * (i - blue[i]);
  cout << ans << "\n";
}
void solve() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i + 1 < n; i++) b[i] = abs(a[i] - a[i + 1]);
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    go(l, r);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  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;
const int MOD = 1e9 + 7;
const int C = 26;
void add(int &x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 1e5 + 10;
long long n, q, a[MAXN], b[MAXN];
int blue[MAXN], red[MAXN], vec[MAXN];
void go(int l, int r) {
  fill(blue + l, blue + r, l - 1);
  fill(red + l, red + r, r);
  int tt = 0;
  for (int i = l; i < r; i++) {
    while (tt && b[vec[tt - 1]] <= b[i]) --tt;
    if (tt) blue[i] = vec[tt - 1];
    vec[tt++] = i;
  }
  tt = 0;
  for (int i = r - 1; i >= l; --i) {
    while (tt && b[vec[tt - 1]] < b[i]) --tt;
    if (tt) red[i] = vec[tt - 1];
    vec[tt++] = i;
  }
  long long ans = 0;
  for (int i = l; i < r; i++) ans += b[i] * (red[i] - i) * (i - blue[i]);
  cout << ans << "\n";
}
void solve() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i + 1 < n; i++) b[i] = abs(a[i] - a[i + 1]);
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    go(l, r);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}

```