題目：
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
5 3
4 7 11 8 10
4 5
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.

題意： 從1開始操作，可以進行兩種操作，第一種是乘以2，第二種是加1。讓你求區間的總共的運算元。

思路： 一個數從1開始操作，要想讓操作次數最少，理所當然是要讓乘以2的操作變多，所以我們可以倒著推一個數所需要的運算元，如果這個數是偶數，我們就讓這個數除以2，然後運算元加1，如果這個數變為奇數了，我們就減1，運算元加1，讓它變成偶數。

程式碼：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=1e5+10;
ll ans[N];
ll solve(ll x){
    int num=0;
    while(x!=1){
        if(x%2==0)x/=2;
        else x--;
        num++;
    }
    return num;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        memset(ans,0,sizeof(ans));
        int n,m,l,r;
        scanf("%d%d",&n,&m);
        ll x;
        for(int i=1;i<=n;i++){
            scanf("%lld",&x);
            ans[i]=ans[i-1]+solve(x);
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d",&l,&r);
            printf("%lld\n",ans[r]-ans[l-1]);
        }
    }
    return 0;
}