#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<long long> vll;
typedef pair<int,int> pint;
typedef pair<long long, long long> pll;

#define MP make_pair
#define PB push_back
#define ALL(s) (s).begin(),(s).end()
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P) 
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P) 
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P) 
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }



bool is_prime(long long n) {
    if (n == 0 || n == 1) return false;
    for (long long i = 2; i*i <= n; ++i) {
        if (n%i ==  0) return false;
    }
    return true;
}



const long long MOD = 4294967311LL;

inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}

inline long long mul(long long a, long long b, long long m) {
    a = mod(a, m); b = mod(b, m);
    if (b == 0) return 0;
    long long res = mul(mod(a + a, m), b>>1, m);
    if (b & 1) res = mod(res + a, m);
    return res;
}

inline long long inv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a/b;
        a = mod(a - t * b, m); swap(a, b);
        u = mod(u - mul(t, v, m), m); swap(u, v);
    }
    return mod(u, m);
}


int N;
int o;
long long Y;

int main() {
    while (cin >> N) {
        long long res = 0;
        
        for (int i = 0; i < N; ++i) {
            cin >> o >> Y;
            
            if (o == 1) res = mod(res + Y, MOD);
            else if (o == 2) res = mod(res - Y, MOD);
            else if (o == 3) res = mul(res, Y, MOD);
            else if (o == 4) res = mul(res, inv(Y, MOD), MOD);
            
            //cout << res << endl;
        }
        
        while (res >= (1LL<<31)) res -= MOD;
        cout << res << endl;
    }
    
    return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

Problem statement

There is a rational number sequence $ X_0, X_1, X_2, ..., X_N $. Each term is defined as follows.


1. $ X_0 = 0 $
2. $ X_i = X_ {i-1} $ $ op_i $ $ Y_i $ ($ 1 \ leq i \ leq N $). However, $ op_i $ is $ + $, $ − $, $ × $, $ ÷ $ Either.



Find $ X_N $.

Constraint

* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq o_i \ leq 4 $
* $ -10 ^ 6 \ leq Y_i \ leq 10 ^ 6 $
* If $ o_i = 4 $, then $ Y_i \ neq 0 $
* $ X_N $ is an integer greater than or equal to $ -2 ^ {31} $ and less than $ 2 ^ {31} $.



input

Input follows the following format. All given numbers are integers.


$ N $
$ o_1 $ $ Y_1 $
$ o_2 $ $ Y_2 $
$ ... $
$ o_N $ $ Y_N $

When $ o_i = 1 $, $ op_i $ is +, when $ o_i = 2 $, $ op_i $ is −, when $ o_i = 3 $, $ op_i $ is ×, and when $ o_i = 4 $, $ op_i $ is ÷.

output

Print the value of $ X_N $ on one line.

Example

Input

4
1 1
4 2
2 4
3 4


Output

-14

### Response

```cpp
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<long long> vll;
typedef pair<int,int> pint;
typedef pair<long long, long long> pll;

#define MP make_pair
#define PB push_back
#define ALL(s) (s).begin(),(s).end()
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P) 
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P) 
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P) 
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }



bool is_prime(long long n) {
    if (n == 0 || n == 1) return false;
    for (long long i = 2; i*i <= n; ++i) {
        if (n%i ==  0) return false;
    }
    return true;
}



const long long MOD = 4294967311LL;

inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}

inline long long mul(long long a, long long b, long long m) {
    a = mod(a, m); b = mod(b, m);
    if (b == 0) return 0;
    long long res = mul(mod(a + a, m), b>>1, m);
    if (b & 1) res = mod(res + a, m);
    return res;
}

inline long long inv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a/b;
        a = mod(a - t * b, m); swap(a, b);
        u = mod(u - mul(t, v, m), m); swap(u, v);
    }
    return mod(u, m);
}


int N;
int o;
long long Y;

int main() {
    while (cin >> N) {
        long long res = 0;
        
        for (int i = 0; i < N; ++i) {
            cin >> o >> Y;
            
            if (o == 1) res = mod(res + Y, MOD);
            else if (o == 2) res = mod(res - Y, MOD);
            else if (o == 3) res = mul(res, Y, MOD);
            else if (o == 4) res = mul(res, inv(Y, MOD), MOD);
            
            //cout << res << endl;
        }
        
        while (res >= (1LL<<31)) res -= MOD;
        cout << res << endl;
    }
    
    return 0;
}

```