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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
template<typename T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
//end

template<unsigned mod=1000000007>struct fp {
   unsigned v;
   static unsigned get_mod(){return mod;}
   unsigned inv() const{
      int tmp,a=v,b=mod,x=1,y=0;
      while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
      if(x<0){x+=mod;} return x;
   }
   fp():v(0){}
   fp(ll x):v(x>=0?x%mod:mod+(x%mod)){}
   fp operator-()const{return fp(-v);}
   fp pow(ll t){fp res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;} return res;}
   fp& operator+=(const fp& x){if((v+=x.v)>=mod)v-=mod; return *this;}
   fp& operator-=(const fp& x){if((v+=mod-x.v)>=mod)v-=mod; return *this;}
   fp& operator*=(const fp& x){v=ll(v)*x.v%mod; return *this;}
   fp& operator/=(const fp& x){v=ll(v)*x.inv()%mod; return *this;}
   fp operator+(const fp& x)const{return fp(*this)+=x;}
   fp operator-(const fp& x)const{return fp(*this)-=x;}
   fp operator*(const fp& x)const{return fp(*this)*=x;}
   fp operator/(const fp& x)const{return fp(*this)/=x;}
   bool operator==(const fp& x)const{return v==x.v;}
   bool operator!=(const fp& x)const{return v!=x.v;}
}; using Fp=fp<>;

Fp dp[5010][2][2],sub[5010][2];

int main(){
   int n,a,b; cin>>n>>a>>b; n+=2;
   if(a>b)swap(a,b);
   sub[0][0]=1;
   rep(i,0,n){
      for(int add=1;i+add<=n;add++)sub[i+add][1]+=sub[i][0];
      for(int add=a;i+add<=n;add++)sub[i+add][0]+=sub[i][1];
   }
   dp[0][0][0]=1;
   rep(i,0,n)rep(k,0,2){
      for(int add=1;i+add<=n and add<a;add++)dp[i+add][0][k]+=dp[i][1][k];
      for(int add=1;i+add<=n;add++){
         int cur=add; if(i==0)cur--; if(i+add==n)cur--;
         dp[i+add][1][cur>=b?1:k]+=dp[i][0][k]*sub[add][1];
      }
   }
   cout<<dp[n][1][1].v<<endl;
   return 0;
}