Snuke has a string x of length N. Initially, every character in x is `0`.

Snuke can do the following two operations any number of times in any order:

* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.



Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).

Constraints

* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N A B


Output

Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).

Examples

Input

4 2 3


Output

11


Input

10 7 2


Output

533


Input

1000 100 10


Output

828178524
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define SZ(x) ((int)x.size())
#define ALL(x) x.begin(),x.end()
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=(a); i<=(b); i++)
#define per(i,a,b) for (register int i=(a); i>=(b); i--)
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> Pii;
typedef pair<ll,ll> Pll;
typedef vector<int> Vi;
template<class T> inline void read(T &x){
	x=0; char c=getchar(); int f=1;
	while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
	while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
template<class T> T gcd(T a, T b){return !b?a:gcd(b,a%b);}
template<class T> inline void umin(T &x, T y){x=x<y?x:y;}
template<class T> inline void umax(T &x, T y){x=x>y?x:y;}
mt19937 R(chrono::system_clock().now().time_since_epoch().count());

const int N = 1<<18|3, mo = 1e9+7, inv2 =(mo+1)/2;
const ll MOD=7LL*mo*mo;
inline void add(int &x, int y){x=x+y<mo?x+y:x+y-mo;}
inline void sub(int &x, int y){x=x-y>=0?x-y:x-y+mo;}
inline int power(int a, int n) {
	int res=1;
	while (n) {
		if (n&1) res=1LL*res*a%mo;
		a=1LL*a*a%mo; n>>=1;
	}
	return res;
}

int fac[N],ifac[N];
void iniC(int n){
	fac[0]=1;rep(i,1,n)fac[i]=1LL*fac[i-1]*i%mo;
	ifac[n]=power(fac[n],mo-2);
	per(i,n-1,0)ifac[i]=1LL*ifac[i+1]*(i+1)%mo;
}
inline int C(int n, int m){return m>=0&&m<=n?1LL*fac[n]*ifac[m]%mo*ifac[n-m]%mo:0;}
int n,a,b,f[2][N],dp[N],dp2[N];

int main() {
	read(n);read(a);read(b);iniC(N-1);
	if(a<b)swap(a,b);//a>=b
	
	f[1][0]=1;
	rep(i,1,n){
		rep(j,0,i-1)add(f[0][i],f[1][j]);
		rep(k,0,i-b)add(f[1][i],f[0][k]);
		dp[i]=(f[0][i])%mo;
		dp2[i]=(f[1][i])%mo;//f[0][i]+
	}
	memset(f,0,sizeof(f));
	
	rep(i,0,1)f[i][0]=1;
	rep(i,1,n){
		rep(k,max(0,i-a+1),i-1){
			add(f[0][i],1ll*f[1][k]*dp[i-k]%mo);
			if(i==n||k==0)add(f[0][i],1ll*f[1][k]*dp2[i-k]%mo);
		}
		rep(k,max(0,i-b+1),i-1)add(f[1][i],f[0][k]);
	}
	int res=((power(2,n)-f[0][n]-f[1][n])%mo+mo)%mo;
	cout<<res;
	return 0;
}
