#pragma region
#pragma GCC target("avx2")
#pragma GCC optimize("03")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std; typedef long double ld; typedef long long ll;
typedef unsigned long long ull;
#define endl "\n"
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define ALL(x) (x).begin(), (x).end()
constexpr int INF=1<<30; constexpr ll LINF=1LL<<60; constexpr ll mod=1e9+7; constexpr int NIL = -1;
template<class T>inline bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>inline bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
#pragma endregion
//-------------------
struct mint {
	ll x;
	mint(ll x=0):x(x%mod){}

	bool operator==(const mint a)const{return x==a.x;}
	bool operator!=(const mint a)const{return x!=a.x;}
	bool operator>=(const mint a){return (x >= a.x)? 1: 0;}
	bool operator<(const mint a){return !(*this>=a);}
	bool operator>(const mint a){return (x > a.x)? 1:0;}
	bool operator<=(const mint a){return !(*this>a);}
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
			return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod-a.x) >= mod) x -= mod;
			return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res+=a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res-=a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res*=a;
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res/=a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t>>1);
		a *= a;  //2 square
		if (t&1) a *= *this; 
		return a;
	}
	// for prime mod
	mint inv() const {
		return pow(mod-2);
	}
};

int depth(int a) {
    int i = 1;
    int cur = 2;
    while(cur-1 < a) {
        i++;
        cur *= 2;
    }
    return i;
}

mint fact[(ll) 1<<19];

mint solve() {
    int d,l,r,k; cin >> d >> l >> r >> k;

    fact[0] = mint(1);
    for(ll i=1; i<=1<<20; i++) fact[i] = fact[i-1]*mint(i);

    int dep_l = depth(l);
    int dep_r = depth(r);
    // cout << dep_l << dep_r << endl;

    int dep_diff = dep_r - dep_l;
    if(k < dep_diff) return mint(0);
    int cnt = 0;
    while(k > dep_diff and cnt <= dep_l - 1) {dep_diff += 2; cnt++;}
    if(k != dep_diff) return mint(0);

    // cout << cnt << endl;
    mint ans;

    int par = dep_l - cnt;


    if(cnt != 0) {
        ans = mint(2).pow(dep_l-par-1) * mint(2).pow(dep_r-par-1) * 2;
        ans *= mint(2).pow(par-1);
        for(int i=2; i<=d; i++) {
            ll num = (1<<(i-1)) - (i==dep_l? 1:0) - (i==dep_r? 1:0);
            ans *= fact[num];
        }
    }
    else {
        ans = mint(2).pow(dep_r - par);
        ans *= mint(2).pow(par-1);
        for(int i=2; i<=d; i++) {
            ll num = (1<<(i-1)) - (i==dep_l? 1:0) - (i==dep_r? 1:0);
            ans *= fact[num];
        }
    } 

    return ans;
}

int main(){
    cin.tie(0); ios::sync_with_stdio(false); //cout << fixed << setprecision(15);
    
    auto res = solve();
    cout << res.x << endl;
    
    return 0;
}