为什么在同一条件变量上使用多个互斥锁会使此代码崩溃？

在下面的代码中，我试图使其最小限度地可验证，运行良好，并做了它应该做的事情(无论我在线程中传递哪个顺序，都按顺序打印1,2,3(。然而，如果我在函数第三行中将m1更改为m2，则此代码将崩溃，并显示消息"terminated without a active exception"。为什么我不能使用相同的条件变量同时锁定两个不同的互斥对象？

#include <functional>
#include <mutex>
#include <condition_variable>
#include <future>
#include <iostream>
void printFirst() {
cout << "1";
}
void printSecond() {
cout << "2";
}
void printThird() {
cout << "3";
}
struct test {
condition_variable c, c2;
int count = 0;
mutex m1,m2;
void first(function<void()> printFirst) {
printFirst();
count++;
c.notify_all();
}
void second(function<void()> printSecond) {
unique_lock<mutex> sL1(m1);
c.wait(sL1,[&]{return count>=1;});
printSecond();
count+=1;
c.notify_all();
}
void third(function<void()> printThird) {
unique_lock<mutex> sL2(m1); //If I make m1, m2, this code crashes
c.wait(sL2,[&]{return count>=2;});
printThird();
}
};
int main() {
test t;
function<void()> printFirstN =[&](){ t.first(printFirst);};
function<void()> printSecondN=[&](){ t.second(printSecond);};
function<void()> printThirdN=[&](){ t.third(printThird);};
std::thread t1(printFirstN);
std::thread t2( printThirdN);
std::thread t3( printSecondN);
t1.join();
t2.join();
t3.join();
}