Codeforces Round #333 (Div. 2) D. Lipshitz Sequence （单调栈）

D. Lipshitz Sequence

time limit per test

memory limit per test

input

output

A function 

 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 

. We'll deal with a more... discrete version of this term.For an array 

, we define it's Lipschitz constant 

 as follows:

• ifn< 2,



• ifn≥ 2,



• over all 1 ≤i<j≤n

In other words, 

 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 

 of size n and q queries of the form [l, r]. For each query, consider the subarray 

; determine the sum of Lipschitz constants of all subarrays of 

.

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 

 and the number of queries respectively.The second line contains n space-separated integers 

 (

).

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 

.

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 

 with length at least 2

The answer to the query is their sum.

题意：给你一组数，q个询问，每次询问一个区间。让你求出这个区间中各个子区间中相邻两个数绝对值的最大值之和。

题解：预处理一下。然后单调栈乱搞一下就可以了。

代码：

#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<cmath>
#include<queue>
#include<set>
#include<stack>
#include <utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)  
#define rep(i,j,k) for (int i = j; i <= k; i++)  
#define per(i,j,k) for (int i = j; i >= k; i--)  
#define lson x << 1, l, mid  
#define rson x << 1 | 1, mid + 1, r  
const int lowbit(int x) { return x&-x; }  
const double eps = 1e-8;  
const int INF = 1e9+7; 
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;  
const ll mod = (1LL<<32);
const int N = 3e5+7; 
const int M=100010;
const int maxn=1e5+7; 
template <class T1, class T2>inline void getmax(T1 &a, T2 b) {if (b>a)a = b;}  
template <class T1, class T2>inline void getmin(T1 &a, T2 b) {if (b<a)a = b;}
int read(){
int v = 0, f = 1;char c =getchar();
while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}
while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();
return v*f;}
int n,q;
int a[maxn],b[maxn];
pii p[maxn];
ll solve(int l,int r)
{
  ll ans=0;
  ll sum=0;
  int rear=0;
  for(int i=l;i<r;i++)
  {
    int num=1;
    while(rear && b[i] >= p[rear].first)
    {
      sum-=(1ll * p[rear].first * p[rear].second);
      num+=p[rear].second;
      rear--;  
    }
    p[++rear].first=b[i];
    p[rear].second=num;
      sum+=(1ll*b[i]*num);
    ans+=sum;
  }
  return ans;
}
int main()
{
  cin>>n>>q;
  for(int i=1;i<=n;i++)cin>>a[i];
  for(int i=1;i<n;i++)b[i]=abs(a[i]-a[i+1]);
  while(q--)
  {
    int l,r;
    cin>>l>>r;
    cout<<solve(l,r)<<endl;
  }
  return 0;
}