Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.
Simon loves fractions very much. Today he wrote out number on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: , where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction.
Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).
The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109).
Print a single number — the answer to the problem modulo 1000000007 (109 + 7).
Examples
2 2
2 2
8
3 3
1 2 3
27
2 2
29 29
73741817
4 5
0 0 0 0
1
Note
In the first sample . Thus, the answer to the problem is 8.
In the second sample, . The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.
In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.
In the fourth sample . Thus, the answer to the problem is 1.
思路是主要的进制的思想
代码:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<set> #include<map> #include<vector> #include<cmath> #define mod 1000000007 const int maxn=1e5+5; typedef long long ll; using namespace std; ll quickpow(ll a,ll b) { ll ans=1; while(b) { if(b&1) ans=(ans*a)%mod; b>>=1; a=(a*a)%mod; } return ans; } ll a[maxn]; int main() { ll n,x; cin>>n>>x; ll s=0; for(int t=0;t<n;t++) { scanf("%lld",&a[t]); s+=a[t]; } for(int t=0;t<n;t++) { a[t]=s-a[t]; } ll ans,cnt=1; sort(a,a+n); a[n]=-1; for(int t=1;t<=n;t++) { if(a[t]!=a[t-1]) { if(cnt%x) { ans=a[t-1]; break; } else { cnt/=x; a[t-1]++; t--; } } else { cnt++; } } ll ss=min(s,ans); cout<<quickpow(x,ss)<<endl; return 0; }