Codeforces Round #157 (Div. 1) C. Little Elephant and LCM (Math, DP)

C. Little Elephant and LCM time limit/test 4 seconds memory limit per test 256 megabytes input standard input output s Tandard output

The Little Elephant loves the LCM (least common multiple) operation of a non-empty set of positive integers. The result of the LCM operation of K positive integers x1,? x2,?...,? x k is the minimum positive integer This is divisible by each of numbers xi.

Let's assume that there is a sequence of integers b1,? b2,?...,? b n. Let ' s denote their lcms as LCM(b1,? b2,?...,? b n) and the maximum of them as Max(b1,? ) b2,?...,? b n). The Little Elephant considers a sequence b good, if LCM(b1,? ) b2,?...,? b n)? =? Max (b1,?) b2,?...,? b n).

The Little Elephant has a sequence of integers a1,? a2,?...,? a n. Help him find the number of good sequences of integers b1,? b2,?...,? b n, such that to all I (1?≤? ) I≤? n) The following condition fulfills:1?≤? b i≤? a Ias the answer can is rather large, print the remainder from dividing it by 1000000007 (109?+?7).

Input

The contains a single positive integer n (1?≤? n. ≤?105)-the number of integers in the sequence a. The second line contains nspace-separated integers a1,? a2,?...,? a n (1?≤? Ai≤?105)-sequence a.

Output

In the "single" Print a single integer-the answer to the problem modulo 1000000007 (109?+?7).

Sample Test (s)
Input

4
1 4 3 2

Output

15

Input

2
6 3

Output

13

Meaning

Give you a sequence of a, find a b sequence, 1?≤?bi?≤?ai, make Max (BI) =LCM (BI), and ask how many of these bi sequences there are.

Ideas:

First to a sort, enumerated i=max (BI), to I factorization, then more than equal to I of the part is very good processing, direct pow_mod () subtraction, less than the part of I arbitrarily take a constraint is enough.

Code:

#include
  
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    #include
    
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       #include 
       
        #include 
        
         #define INF 0x3f3f3f3f #define MAXN 100005 #define MOD 1000000007 typedef lon G Long ll; using namespace Std; int n; int A[MAXN]; ll Pow_mod (ll x,ll N) {ll res = 1 while (n) {if (n&1) res = res * x%mod; x = x * x%mod; n >>= 1;} return res ; } void Solve () {int i,j; ll ans=0,res; sort (a+1,a+n+1); for (i=1;i<=a[n];i++)//enum answer {vector 
         
          fac; for (j=1 ; j*j<=i;j++)//Factor {if (i%j==0) {fac.push_back (j); if (j*j!=i) fac.push_back (i/j);} sort (Fac.begin (), Fac.end ()) ; int pos,pre=1; Res=1; For (j=1;j 
          
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