#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
inline void solve(), read();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
long long n, q;
vector<long long> a, d;
vector<long long> posl, posr;
void getmin() {
  posl.resize(n - 1, -1);
  stack<long long> st;
  for (long long i = 0; i < n - 1; i++) {
    while (!st.empty() && d[st.top()] < d[i]) {
      st.pop();
    }
    if (!st.empty()) {
      posl[i] = st.top();
    }
    st.push(i);
  }
}
void getmax() {
  posr.resize(n - 1, n);
  stack<long long> st;
  for (long long i = n - 2; i >= 0; i--) {
    while (!st.empty() && d[st.top()] <= d[i]) {
      st.pop();
    }
    if (!st.empty()) {
      posr[i] = st.top();
    }
    st.push(i);
  }
}
long long query(long long l, long long r) {
  long long ans = 0;
  for (long long i = l; i <= r; i++) {
    long long l1 = max(l - 1, posl[i]);
    long long r1 = min(posr[i], r + 1);
    ans += (i - l1) * (r1 - i) * d[i];
  }
  return ans;
}
inline void solve() {
  d.resize(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    d[i] = abs(a[i + 1] - a[i]);
  }
  getmin(), getmax();
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    l -= 1, r -= 2;
    long long cur = query(l, r);
    cout << cur << "\n";
  }
}
inline void read() {
  cin >> n >> q;
  a.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
}

### 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
inline void solve(), read();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
long long n, q;
vector<long long> a, d;
vector<long long> posl, posr;
void getmin() {
  posl.resize(n - 1, -1);
  stack<long long> st;
  for (long long i = 0; i < n - 1; i++) {
    while (!st.empty() && d[st.top()] < d[i]) {
      st.pop();
    }
    if (!st.empty()) {
      posl[i] = st.top();
    }
    st.push(i);
  }
}
void getmax() {
  posr.resize(n - 1, n);
  stack<long long> st;
  for (long long i = n - 2; i >= 0; i--) {
    while (!st.empty() && d[st.top()] <= d[i]) {
      st.pop();
    }
    if (!st.empty()) {
      posr[i] = st.top();
    }
    st.push(i);
  }
}
long long query(long long l, long long r) {
  long long ans = 0;
  for (long long i = l; i <= r; i++) {
    long long l1 = max(l - 1, posl[i]);
    long long r1 = min(posr[i], r + 1);
    ans += (i - l1) * (r1 - i) * d[i];
  }
  return ans;
}
inline void solve() {
  d.resize(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    d[i] = abs(a[i + 1] - a[i]);
  }
  getmin(), getmax();
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    l -= 1, r -= 2;
    long long cur = query(l, r);
    cout << cur << "\n";
  }
}
inline void read() {
  cin >> n >> q;
  a.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
}

```