考虑前两个,如果两个一样,那么就是m^(n-1),不一样的话就是f(n-1),
式子:f(n)=f(n-1)*(m-1)+m^(n-1),
不难凑出:(f(n)-m^n)=(f(n-1)-m^(n-1))(m-1).
f(2)为m.
#include<bits/stdc++.h>
using namespace std;#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define ll long long#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define root l,r,rt
#define mst(a,b) memset((a),(b),sizeof(a))
#define pii pair<int,int>
#define fi first
#define se second
#define mk(x,y) make_pair(x,y)
const int mod=1e5+3;
const int maxn=1e5+100;
const int ub=1e6;
ll powmod(ll x,ll y){x%=mod,y%=(mod-1);ll t;for(t=1;y;y>>=1,x=x*x%mod) if(y&1) t=t*x%mod;return t;
}
ll gcd(ll x,ll y){if(y==0) return x;return gcd(y,x%y);
}
ll n,m;
int main(){cin>>m>>n;ll tmp=(powmod(m,n-1)-powmod(m-1,n-1)+mod);cout<<m%mod*tmp%mod<<endl;return 0;
}