// Original C++ solution (slower):
#include <iostream>

#include <limits>

#include <cfenv>

#include <cmath>

#include <algorithm>

#include <array>

#include <bitset>

#include <cstring>

#include <map>

#include <stack>

#include <set>

#include <tuple>

#include <queue>

#include <vector>

#include <cmath>

#include <random>

#include <math.h>

#include <list>

#include <random>

#include <functional>





#define FOR(i, a, b) for(int (i) = (a); (i) < (b); ++(i))

#define REP(i, n) FOR(i, 0, n)

#define rREP(i, n) for(int (i) = (n) - 1; (i) >= 0; --(i))

#define ALL(TheArray) TheArray.begin(), TheArray.end()



using lli = long long int;

using pii = std::pair<int, int>;



template <class T> inline bool chmax(T& a, T b){

    if(a < b){a = b; return true;}

    return false;

}

template <class T> inline bool chmin(T& a, T b){

    if(a > b){a = b; return true;}

    return false;

}





// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~





constexpr int N = 1000;

constexpr int mod = 1e9 + 7;

std::array<std::array<lli, N + 1>, N + 1> A;

std::array<lli, N+1> F, invF;



struct faster{

    lli Power[N+1], invP[N+1];

    constexpr faster(): Power(), invP(){

        Power[0] = 1;

        for(int i = 1; i <= N; ++i) Power[i] = (Power[i-1] * i) % mod;

        invP[N] = inverse(Power[N]);

        for(int i = N; i > 0; --i) invP[i-1] = (invP[i] * i) % mod;

    }

    constexpr void swap(lli &a, lli &b) const {

        a ^= b; b ^= a; a ^= b;

    }

    constexpr lli inverse(lli a) const noexcept{

        lli b = mod, u = 1, v = 0;

        while (b != 0) { 

            lli t = a / b; a -= t * b; swap(a, b);

            u -= t * v; swap(u, v);

        }

        u %= mod; if (u < 0) u += mod;

        return u;

    }

};



// 累乗の余りを求める関数

inline lli power(lli a, int n, lli mod_number___ = mod){

    lli res = 1, waiting = a;

    while(n){ if(n & 1){ 

        (res *= waiting) %= mod_number___;}

        (waiting *= waiting) %= mod_number___; 

        n >>= 1;

    }

    return res;

}



/* n人でp人グループを k 個作る分け方は

    nCpk * pkCp * (k-1)pCp * ... * pCp / k!

    = {n!/(n-pk)!pk!} * {pk!/(k-1)p!p!} * {(p-1)k!/(k-2)p!p!} * ... * pCp / k!

    = {n!/(n-pk)!pk!} * pk! / (p!)^k / k!

    = n! / { (n - pk)! * (p!)^k * k! }

*/



constexpr faster Init;



int main(void){

    int n, a, b, c, d; scanf("%d%d%d%d%d", &n, &a, &b, &c, &d);



    int bMa = b - a;

    A[0][0] = 1;

    for(int i = 0; i <= bMa; ++i){

        int p = a + i;

        for(int j = 0; j <= n; ++j){

            (A[i+1][j] += A[i][j]) %= mod;

            lli C = power(Init.invP[p], c) * Init.invP[j] % mod;

            for(int k = c; k <= d and j + k * p <= n; ++k){

                lli c = (Init.Power[j + k * p] * Init.invP[k] % mod) * C % mod;

                (A[i + 1][j + k * p] += c * A[i][j]) %= mod;

                (C *= Init.invP[p]) %= mod;

            }

        }

    }



    printf("%lld\n", A[bMa+1][n]);

    



    











    return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <algorithm>

#include <array>





#define FOR(i, a, b) for(int (i) = (a); (i) < (b); ++(i))

#define REP(i, n) FOR(i, 0, n)



using lli = long long int;



template <class T> inline bool chmax(T& a, T b){

    if(a < b){a = b; return true;}

    return false;

}

template <class T> inline bool chmin(T& a, T b){

    if(a > b){a = b; return true;}

    return false;

}





// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~





constexpr int N = 1000;

constexpr int mod = 1e9 + 7;

std::array<std::array<lli, N + 1>, 2> A;



struct faster{

    lli Power[N+1], invP[N+1];

    constexpr faster(): Power(), invP(){

        Power[0] = 1;

        for(int i = 1; i <= N; ++i) Power[i] = (Power[i-1] * i) % mod;

        invP[N] = inverse(Power[N]);

        for(int i = N; i > 0; --i) invP[i-1] = (invP[i] * i) % mod;

    }

    constexpr void swap(lli &a, lli &b) const {

        a ^= b; b ^= a; a ^= b;

    }

    constexpr lli inverse(lli a) const noexcept{

        lli b = mod, u = 1, v = 0;

        while (b != 0) { 

            lli t = a / b; a -= t * b; swap(a, b);

            u -= t * v; swap(u, v);

        }

        u %= mod; if (u < 0) u += mod;

        return u;

    }

};



// 累乗の余りを求める関数

inline lli power(lli a, int n, lli mod_number___ = mod){

    lli res = 1, waiting = a;

    while(n){ if(n & 1){ 

        (res *= waiting) %= mod_number___;}

        (waiting *= waiting) %= mod_number___; 

        n >>= 1;

    }

    return res;

}



constexpr faster Init;



int main(void){

    int n, a, b, c, d; scanf("%d%d%d%d%d", &n, &a, &b, &c, &d);



    A[0][0] = 1;

    for(int p = a; p <= b; ++p){

        for(int j = 0; j <= n; ++j) A[1][j] = A[0][j];

        for(int j = 0; j <= n; ++j){

            lli C = power(Init.invP[p], c) * Init.invP[j] % mod;

            int ptr = j + c * p; 

            int e = (n - j) / p; if(e > d) e = d;

            for(int k = c; k <= e; ++k, ptr += p){

                lli c = (Init.Power[ptr] * Init.invP[k] % mod) * C % mod;

                (A[1][ptr] += c * A[0][j]) %= mod;

                (C *= Init.invP[p]) %= mod;

            }

        }

        std::swap(A[0], A[1]);

    }

    printf("%lld\n", A[0][n]);

    return 0;

}

// Fastest known solution for this problem: