print(eval(input()))

print(eval(input()))

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

const int MAXN = 200000;

template <typename T>
struct Stack{
	T arr[MAXN];
	int now;
	Stack() : now(0) {}
	T top() {
		return arr[now-1];
	}
	void pop() {
		now--;
	}
	void push(T val) {
		arr[now++] = val;
	}
	T size() {
		return now;
	}
};

template <typename T>
struct Queue{
	T arr[MAXN];
	int head, tail;
	Queue() : head(0), tail(0) {}
	int front() {
		return arr[head];
	}
	void pop() {
		head++;
		if (head == MAXN) head = 0;
	}
	void push(int val) {
		arr[tail++] = val;
		if (tail == MAXN) tail = 0;
	}
	int size() {
		return (tail + MAXN - head) % MAXN;
	}
};

string str;
int num, k=1;
Stack <int> stk_num; // 代表數字的 stack
Stack <char> stk_sym; // 代表運算符號的 stack
Queue <int> que_num; // 代表數字的 queue
Queue <char> que_sym; // 代表運算符號的 queue

void calc() {
	int a = stk_num.top(), b;
	stk_num.pop();
	char op;
	while (stk_sym.size() && stk_sym.top() != ')') {
		b = stk_num.top();
		stk_num.pop();
		op = stk_sym.top();
		stk_sym.pop();
		if (op == '+' || op == '-') { // 遇到加減運算先不處理
			que_sym.push(op);
			que_num.push(a);
			a = b;
		} else if (op == '*') {
			a = a * b;
		} else if (op == '/') {
			a = a / b;
		}
	}
	que_num.push(a);
	// 這邊回來解決加減運算
	a = que_num.front();
	que_num.pop();
	while (que_sym.size()) {
		b = que_num.front();
		que_num.pop();
		op = que_sym.front();
		que_sym.pop();
		if (op == '+') {
			a = a + b;
		} else if (op == '-') {
			a = a - b;
		}
	}
	if (stk_sym.size()) stk_sym.pop(); // 若並非結尾，則將 ')' pop 出來
	stk_num.push(a); // 將運算結果放回 stack 中
}

signed main() {
	cin >> str;
	for (int i = str.size()-1; i >= 0; i--) { // 由後往前！
		if (str[i] >= '0' && str[i] <= '9') {
			num = num + k * (str[i] - '0');
			if (i-1 >= 0 && str[i-1] >= '0' && str[i-1] <= '9') {
				k = k * 10; // 如果下一位還是數字的話
			} else {
				stk_num.push(num);
				num = 0, k = 1;
			}
		} else if (str[i] != '(') {
			stk_sym.push(str[i]); // 遇到正常運算符號就 push 進 stack
		} else if (str[i] == '(') {
			calc(); // 遇到 '(' 就回推
		}
	}
	calc(); // 結尾回推
	cout << stk_num.top() << '\n';
	return 0;
}

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

const int MAXN = 1000000;

struct Deque{
	int arr[MAXN], head, tail;
	Deque() : head(0), tail(0) {}
	int front() {
		return arr[head];
	}
	int back() {
		if (tail == 0) return arr[MAXN-1];
		return arr[tail-1];
	}
	void pop_front() {
		head++;
		if (head == MAXN) head = 0;
	}
	void pop_back() {
		if (tail == 0) tail = MAXN;
		tail--;
	}
	void push_front(int val) {
		if (head == 0) head = MAXN;
		arr[--head] = val;
	}
	void push_back(int val) {
		arr[tail++] = val;
		if (tail == MAXN) tail = 0;
	}
	int size() {
		return (tail + MAXN - head) % MAXN;
	}
} deq;

int N, K;
int x[MAXN+1];

signed main() {
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> N >> K;
	for (int i = 1; i <= N; i++) {
		cin >> x[i];
	}
	// 先處理前 K-1 個
	for (int i = 1; i <= K-1; i++) {
		while (deq.size() && x[i] >= x[deq.back()]) {
			deq.pop_back();
		}
		deq.push_back(i);
	}
	for (int i = K; i <= N; i++) {
		if (deq.size() && i - deq.front() >= K) deq.pop_front();
		// 如果距離和自己超過 K-1 就代表不在這次區間的考慮範圍

		while (deq.size() && x[i] >= x[deq.back()]) {
			deq.pop_back();
		}
		deq.push_back(i);
		cout << x[deq.front()] << " \n"[i==N];
		// 當前的 front() 即是這段區間的最大值
	}
	return 0;
}

#include <bits/stdc++.h>
#define pii pair<int,int>
#define ff first
#define ss second
using namespace std;

const int MAXN = 200000;
const int INF = 1e9;
int N, L;
int x, k;
int arr[MAXN+1];
long long ans;
set <pii> S;

signed main() {
	cin >> N >> L;
	S.insert(pii(0, L));
	for (int i = 1; i <= N; i++) {
		cin >> x >> k;
		arr[k] = x;
	}
	for (int i = 1; i <= N; i++) {
		pii pr = *prev(S.lower_bound(pii(arr[i], INF)));
		ans += pr.ss - pr.ff;
		if (arr[i] == 0 || arr[i] == L) continue;
		S.erase(pr);
		S.insert(pii(pr.ff, arr[i]));
		S.insert(pii(arr[i], pr.ss));
	}
	cout << ans << '\n';
	return 0;
}