#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > matmul(vector<vector<long long> > &a,
                                  vector<vector<long long> > &b) {
  int i, j, k, n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      long long temp = 0;
      for (k = 0; k < n; k++)
        temp = (temp + a[i][k] * b[k][j]) % (1000000007LL);
      ans[i][j] = temp;
    }
  return ans;
}
vector<vector<long long> > matpwr(vector<vector<long long> > &a, long long p) {
  int n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (int i = 0; i < n; i++) ans[i][i] = 1;
  while (p) {
    if (p & 1) ans = matmul(ans, a);
    a = matmul(a, a);
    p >>= 1;
  }
  return ans;
}
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long invMod(long long n, long long mod = (1000000007LL)) {
  return pwr(n, mod - 2);
}
int n, arr[100005];
int main() {
  ios_base::sync_with_stdio(0);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  while (q--) {
    int a, b;
    cin >> a >> b;
    long long ans = 0, temp = 0;
    stack<pair<long long, long long> > st;
    for (int i = a + 1; i <= b; i++) {
      long long val = abs(arr[i] - arr[i - 1]);
      long long cnt = 1;
      while (!st.empty() && st.top().first < val) {
        cnt += st.top().second;
        temp -= st.top().first * st.top().second;
        st.pop();
      }
      st.push(pair<long long, long long>(val, cnt));
      temp += val * cnt;
      ans += temp;
    }
    cout << ans << endl;
  }
  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;
vector<vector<long long> > matmul(vector<vector<long long> > &a,
                                  vector<vector<long long> > &b) {
  int i, j, k, n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      long long temp = 0;
      for (k = 0; k < n; k++)
        temp = (temp + a[i][k] * b[k][j]) % (1000000007LL);
      ans[i][j] = temp;
    }
  return ans;
}
vector<vector<long long> > matpwr(vector<vector<long long> > &a, long long p) {
  int n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (int i = 0; i < n; i++) ans[i][i] = 1;
  while (p) {
    if (p & 1) ans = matmul(ans, a);
    a = matmul(a, a);
    p >>= 1;
  }
  return ans;
}
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long invMod(long long n, long long mod = (1000000007LL)) {
  return pwr(n, mod - 2);
}
int n, arr[100005];
int main() {
  ios_base::sync_with_stdio(0);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  while (q--) {
    int a, b;
    cin >> a >> b;
    long long ans = 0, temp = 0;
    stack<pair<long long, long long> > st;
    for (int i = a + 1; i <= b; i++) {
      long long val = abs(arr[i] - arr[i - 1]);
      long long cnt = 1;
      while (!st.empty() && st.top().first < val) {
        cnt += st.top().second;
        temp -= st.top().first * st.top().second;
        st.pop();
      }
      st.push(pair<long long, long long>(val, cnt));
      temp += val * cnt;
      ans += temp;
    }
    cout << ans << endl;
  }
  return 0;
}

```