题目：

https://ac.nowcoder.com/acm/contest/23479/J

求[l,r]区间内所有合数的最小公倍数%（1e9+7）

  1. 用埃氏筛筛掉所有质数，剩下为合数

  2. 求最小公倍数有3种方法

    • 直接求最小公倍数
    • 用$A*B/gcd(A,B)$就是两数之积 / 最大公因数
    • 分解质因数：$A = 48=22223$ $B=210=2357$，然后两者共有的质因数C:（2，3），两者不共有的A:（2，2，2）,B:（5，7）三者相乘 $232225*7 = 1680$

3.每一个合数的质因子都用unordered_map来维护，看这个质因子是否多于之前质因子的个数，将所有维护好的质因子相乘就是最小公倍数

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#define debug(a) cout << #a << " = " << a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<' '<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<' '<<#b<<" = "<<b<<' '<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<' '<<#b<<" = "<<b<<' '<<#c<<" = "<<c<<' '<<#d<<" = "<<d<<endl;
#define debugx(a, idx) cout << #a << "[" << idx << "] = " << a[idx] << endl;
#define debugarr(a, x) cout<<#a<<": "; for(int i = 0; i < x; i++) cout << a[i] << ' '; cout << endl;
#define debugvec(a) cout<<#a<<": "; for(int i = 0; i < a.size(); i++) cout << a[i] << ' '; cout << endl;
#define sc(x)  scanf_s("%d",&x)
#define SC(x)  scanf_s("%lld",&x)

#include <iostream>
#include<algorithm>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pair<int, int>> piii;
const int N = 30010;

//快速幂
//ll qpow(ll b, ll n) {
//	ll ret = 1;
//	b %= mod;
//	for (; n; n >>= 1, b = (b * b) % mod)
//		if (n & 1) ret = (ret * b) % mod;
//	return ret;
//}
ll res = 1;
const int mod = 1000000007;
bool st[N];
unordered_map<int, bool> mp;
unordered_map<int, int> mp2;
ll gcd(ll a, ll b)
{
	ll temp = 0;
	while (temp = a % b)
	{
		a = b;
		b = temp;
	}
	return b;
}

void divide(int n) {
	unordered_map<int, int> mp1;
	mp1.clear();
	for (int i = 2; i <= n / i; i++) {
		if (n % i == 0) {
			int s = 0;
			while (n % i == 0) {
				n /= i;
				s++;
			}
			//cout << i << " " << s << endl;
			mp1[i] += s;
		}
	}
	//if (n > 1)cout << n << " " << 1 << endl;
	//puts("");
	if(n > 1)mp1[n] += 1;
	for (auto m : mp1) {
		int x = m.first, y = m.second;
		//cout << x << " " << y << endl;
		if (mp2[x] < y)for (int i = 0; i < y - mp2[x]; i++)res = (res % mod * x) % mod;
		mp2[x] = max(y, mp2[x]);
	}
	//cout << endl;
}

void get_prime(int n) {
	for (int i = 2; i <= n; i++) {
		if (!st[i])mp[i] = true;//把素数存起来
		for (int j = i; j <= n; j += i) {//不管是合数还是质数，都用来筛掉后面它的倍数
			st[j] = true;
		}
	}
}

void f() {
	int l, r;
	cin >> l >> r;
	get_prime(r);
	vector<int> v;
	for (int i = l; i <= r; i++) {
		if (mp[i] == false) {
			v.push_back(i);
		}
	}
	int n = v.size();
	if (n == 0) {
		cout << -1 << endl;
		return;
	}
	//res = v[0];
	//a * b / gcd(a, b) << endl;
	//res = ((res * v[i]) % mod / gcd(res, v[i])) % mod;
	for (int i = 0; i < n; i++) {
		divide(v[i]);
	}
	cout << res % mod<< endl;
}
int main()
{
	f();
	return 0;
}