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

Input
2 2
2 2
Output
8
Input
3 3
1 2 3
Output
27
Input
2 2
29 29
Output
73741817
Input
4 5
0 0 0 0
Output
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+;
typedef long long ll;
using namespace std;
ll quickpow(ll a,ll b)
{
ll ans=;
while(b)
{
if(b&)
ans=(ans*a)%mod;
b>>=;
a=(a*a)%mod;
}
return ans;
}
ll a[maxn];
int main()
{
ll n,x;
cin>>n>>x;
ll s=;
for(int t=;t<n;t++)
{
scanf("%lld",&a[t]);
s+=a[t];
}
for(int t=;t<n;t++)
{
a[t]=s-a[t];
}
ll ans,cnt=;
sort(a,a+n);
a[n]=-;
for(int t=;t<=n;t++)
{
if(a[t]!=a[t-])
{
if(cnt%x)
{
ans=a[t-];
break;
}
else
{
cnt/=x;
a[t-]++;
t--;
}
}
else
{
cnt++;
}
}
ll ss=min(s,ans);
cout<<quickpow(x,ss)<<endl;
return ;
}

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