#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f, mod = 1e9 + 7;
const int maxn = 1e5 + 1000;
const long double PI = acos(-1.0);
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
int n, q;
long long a[maxn], st[maxn], le[maxn], re[maxn], g[maxn];
long long stt(int l, int r) {
  memset(le, 0, sizeof(le));
  memset(re, 0, sizeof(re));
  int top = -1;
  for (int i = l; i <= r; i++) {
    le[i] = 1;
    while (a[i] >= st[top] && top != -1) {
      le[i] += le[g[top]];
      top--;
    }
    st[++top] = a[i];
    g[top] = i;
  }
  top = -1;
  for (int i = r; i >= l; i--) {
    re[i] = 1;
    while (a[i] > st[top] && top != -1) {
      re[i] += re[g[top]];
      top--;
    }
    st[++top] = a[i];
    g[top] = i;
  }
  long long sum = 0;
  for (int i = l; i <= r; i++) sum += a[i] * (le[i]) * (re[i]);
  return sum;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i - 1] = abs(a[i] - a[i - 1]);
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    cout << stt(l, r - 1) << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 inf = 0x3f3f3f3f, mod = 1e9 + 7;
const int maxn = 1e5 + 1000;
const long double PI = acos(-1.0);
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
int n, q;
long long a[maxn], st[maxn], le[maxn], re[maxn], g[maxn];
long long stt(int l, int r) {
  memset(le, 0, sizeof(le));
  memset(re, 0, sizeof(re));
  int top = -1;
  for (int i = l; i <= r; i++) {
    le[i] = 1;
    while (a[i] >= st[top] && top != -1) {
      le[i] += le[g[top]];
      top--;
    }
    st[++top] = a[i];
    g[top] = i;
  }
  top = -1;
  for (int i = r; i >= l; i--) {
    re[i] = 1;
    while (a[i] > st[top] && top != -1) {
      re[i] += re[g[top]];
      top--;
    }
    st[++top] = a[i];
    g[top] = i;
  }
  long long sum = 0;
  for (int i = l; i <= r; i++) sum += a[i] * (le[i]) * (re[i]);
  return sum;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i - 1] = abs(a[i] - a[i - 1]);
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    cout << stt(l, r - 1) << endl;
  }
  return 0;
}

```