题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4602
把n等效为排成一列的n个点，然后就是取出其中连续的k个点。分两种情况，一种是不包含两端，2^( n−k−2 ) ∗ (n−k−1) ，另一种是包含两端：2 ∗ 2^( n – k − 1)。然后特殊情况特判一下。。
//STATUS:C++_AC_31MS_248KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=;
const LL INF=0x3f3f3f3f;
const int MOD=,STA=;
const LL LNF=1LL<<;
const double EPS=1e-;
const double OO=1e15;
const int dx[]={-,,,};
const int dy[]={,,,-};
const int day[]={,,,,,,,,,,,,};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End int T,n,m; LL Pow(LL n,int m)
{
LL ret=;
for(;m;m>>=){
if(m&)ret=(ret*n)%MOD;
n=(n*n)%MOD;
}
return ret;
} int main()
{
// freopen("in.txt","r",stdin);
int i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
if(m>n)
printf("0\n");
else if(n==m)
printf("1\n");
else
printf("%I64d\n",(Pow(,n-m)+(m<n-?(n-m-)*Pow(,n-m-):))%MOD);
}
return ;
}