#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
const int maxm = 20;
long long n, q;
long long data[maxn];
long long d[maxn][maxm];
void RMQ_init(long long len) {
  long long x, y;
  for (long long i = 0; i < len; ++i) {
    d[i][0] = i;
  }
  for (long long j = 1; (1 << j) <= len; ++j) {
    for (long long i = 0; i + (1 << j) - 1 < len; ++i) {
      x = d[i][j - 1];
      y = d[i + (1 << (j - 1))][j - 1];
      if (data[x] > data[y])
        d[i][j] = x;
      else
        d[i][j] = y;
    }
  }
}
long long RMQ(long long L, long long R) {
  long long k = 0;
  long long len = R - L + 1;
  long long x, y;
  while ((1 << (k + 1)) <= len) {
    ++k;
  }
  x = d[L][k];
  y = d[R - (1 << k) + 1][k];
  if (data[x] > data[y])
    return x;
  else
    return y;
}
long long query(long long L, long long R) {
  long long ans = 0;
  long long val = 0, index = 0;
  if (L > R) return ans;
  index = RMQ(L, R);
  long long cnt[2] = {0, 0};
  if (index >= L) cnt[0] = (index - L + 1);
  if (R >= index) cnt[1] = (R - index + 1);
  ans += data[index] * cnt[0] * cnt[1];
  return ans + query(L, index - 1) + query(index + 1, R);
}
int main() {
  while (cin >> n >> q) {
    for (long long i = 0; i < n; ++i) {
      cin >> data[i];
    }
    for (long long i = 0; i < n - 1; ++i) {
      data[i] = abs(data[i + 1] - data[i]);
    }
    RMQ_init(n - 1);
    long long tmp = RMQ(4, 7);
    long long L, R;
    for (long long i = 0; i < q; ++i) {
      cin >> L >> R;
      --L;
      --R;
      cout << query(L, R - 1) << endl;
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 maxn = 100000 + 10;
const int maxm = 20;
long long n, q;
long long data[maxn];
long long d[maxn][maxm];
void RMQ_init(long long len) {
  long long x, y;
  for (long long i = 0; i < len; ++i) {
    d[i][0] = i;
  }
  for (long long j = 1; (1 << j) <= len; ++j) {
    for (long long i = 0; i + (1 << j) - 1 < len; ++i) {
      x = d[i][j - 1];
      y = d[i + (1 << (j - 1))][j - 1];
      if (data[x] > data[y])
        d[i][j] = x;
      else
        d[i][j] = y;
    }
  }
}
long long RMQ(long long L, long long R) {
  long long k = 0;
  long long len = R - L + 1;
  long long x, y;
  while ((1 << (k + 1)) <= len) {
    ++k;
  }
  x = d[L][k];
  y = d[R - (1 << k) + 1][k];
  if (data[x] > data[y])
    return x;
  else
    return y;
}
long long query(long long L, long long R) {
  long long ans = 0;
  long long val = 0, index = 0;
  if (L > R) return ans;
  index = RMQ(L, R);
  long long cnt[2] = {0, 0};
  if (index >= L) cnt[0] = (index - L + 1);
  if (R >= index) cnt[1] = (R - index + 1);
  ans += data[index] * cnt[0] * cnt[1];
  return ans + query(L, index - 1) + query(index + 1, R);
}
int main() {
  while (cin >> n >> q) {
    for (long long i = 0; i < n; ++i) {
      cin >> data[i];
    }
    for (long long i = 0; i < n - 1; ++i) {
      data[i] = abs(data[i + 1] - data[i]);
    }
    RMQ_init(n - 1);
    long long tmp = RMQ(4, 7);
    long long L, R;
    for (long long i = 0; i < q; ++i) {
      cin >> L >> R;
      --L;
      --R;
      cout << query(L, R - 1) << endl;
    }
  }
  return 0;
}

```