極致的 overflow 題目，利用四角錐數再用 BS 找 sqrt
//
// GGGGGGGGGGGGG CCCCCCCCCCCCC AAA
// GGG::::::::::::G CCC::::::::::::C A:::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::::A
// G:::::G GGGGGG C:::::C CCCCCC A:::::::::A
//G:::::G C:::::C A:::::A:::::A
//G:::::G C:::::C A:::::A A:::::A
//G:::::G GGGGGGGGGGC:::::C A:::::A A:::::A
//G:::::G G::::::::GC:::::C A:::::A A:::::A
//G:::::G GGGGG::::GC:::::C A:::::AAAAAAAAA:::::A
//G:::::G G::::GC:::::C A:::::::::::::::::::::A
// G:::::G G::::G C:::::C CCCCCC A:::::AAAAAAAAAAAAA:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::A A:::::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A A:::::A
// GGG::::::GGG:::G CCC::::::::::::C A:::::A A:::::A
// GGGGGG GGGG CCCCCCCCCCCCCAAAAAAA AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef unsigned long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 30005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
clock\_t start\_time=clock(), end\_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
Int n;
Int mod=100000000;
Int bs(Int k){
Int l=0,r=3037000499LL;
while(r>=l){
Int m=(r+l)/2;
// debug("%lld\\n",m);
if(m\*m>k)r=m-1;
else l=m+1;
}
return r;
}
void solve(){
if(n%2==0)n--;
Int rt=bs(n);
Int t=rt-1;
Int ans=1LL;
Int m\[3\]={t,t+1,2\*t+1};
for(int i=0,fg1=0,fg2=0;i<3;i++){
if(m\[i\]%2==0&&!fg1){
m\[i\]/=2;
fg1=1;
}
if(m\[i\]%3==0&&!fg2){
m\[i\]/=3;
fg2=1;
}
}
// Int ans2=(t\*(t+1)\*(2\*t+1))/6;
// ans2%=mod;
// debug("%llu %llu %llu\\n",ans,ans2,t);
ans=(ans\*m\[0\])%mod;
ans=(ans\*m\[1\])%mod;
ans=(ans\*m\[2\])%mod;
Int nowdigit=t\*(t+1)/2;
Int x;
if(t&1LL){
x=t/2+1;
nowdigit+=t/2+1;
}else{
x=t/2;
nowdigit+=t/2;
}
x=(x\*x);
ans=(ans+x%mod)%mod;
// debug("%lld\\n",ans);
Int limit=nowdigit\*2-1;
Int num=(n-limit)/2;
ans=(ans+((num%mod)\*rt)%mod)%mod;
// debug("%lld %lld\\n",num,rt);
// debug("%lld %lld %lld\\n",limit,num,x);
printf("%lld\\n",ans);
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%llu",&n)&&n){
solve();
}
}