F - Building Numbers Gym - 101502F

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本文介绍了一个算法问题，旨在寻找构建特定数列范围内所有数字所需的最小操作总数。通过定义的操作集（加1和加倍），文章提供了一个示例和解决方案，展示了如何有效地计算从1开始构建一系列数字所需的最少步骤。

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In this problem, you can build a new number starting from 1, by performing the following operations as much as you need:

Add 1 to the current number.
Add the current number to itself (i.e. multiply it by 2).
For example, you can build number 8 starting from 1 with three operations . Also, you can build number 10 starting from 1 with five operations .

You are given an array a consisting of n integers, and q queries. Each query consisting of two integers l and r, such that the answer of each query is the total number of operations you need to preform to build all the numbers in the range from l to r (inclusive) from array a, such that each number ai (l ≤ i ≤ r) will be built with the minimum number of operations.

Input

The first line contains an integer T (1 ≤ T ≤ 50), where T is the number of test cases.

The first line of each test case contains two integers n and q (1 ≤ n, q ≤ 105), where n is the size of the given array, and q is the number of queries.

The second line of each test case contains n integers a1, a2, …, an (1 ≤ ai ≤ 1018), giving the array a.

Then q lines follow, each line contains two integers l and r (1 ≤ l ≤ r ≤ n), giving the queries.

Output

For each query, print a single line containing its answer.

Example

Input

1 //数据组数
5 3 // 数组中数据的个数，几次查找访问
4 7 11 8 10 //5个数据，求达到每个数据所需要的最短步数
4 5 三次访问，代表达到数组中第四个数据到第五个数据，所需要的步数总和，最后输出
1 5
3 3

Output

7
18
5

Note

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

In the first query, you need 3 operations to build number 8, and 4 operations to build number 10. So, the total number of operations is 7.

题目大意：

见上面样例的详细解释

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll f(ll x)
{
  ll bns=0;
  while(x>0)
  {
    if(x%2==0) x=x/2;
    else x--;
    bns++;
  }
  return bns-1;
}

int main()
{
   ll t,i,m,n,l,r;
   ll a[100005];
   ll bns[100005];
   scanf("%lld",&t);
   while(t--)
   {
     memset(bns,0,sizeof(bns));
     memset(a, 0, sizeof(a));
     scanf("%lld %lld",&n,&m);
     for(i=0;i<n;i++)
     {
       scanf("%lld",&a[i]);
       bns[i] = f(a[i]);
     }
     for(i = 1; i < n; i++)
     {
         bns[i] = bns[i-1] + f(a[i]);          //!求前缀和 （前i项的步数和）
     }
     for(i=1;i<=m;i++)
     {
       scanf("%lld %lld",&l,&r);
       printf("%lld\n",bns[r-1]-bns[l-2]);//数组从从0开始（为什么i-2 可以自己找纸算算）
     }
   }
   return 0;