// C++ code to convert infix expression to postfix

#include <bits/stdc++.h>

using namespace std;

// Function to return precedence of operators

int prec(char c)

{

    if (c == '^')

        return 3;

    else if (c == '/' || c == '*')

        return 2;

    else if (c == '+' || c == '-')

        return 1;

    else

        return -1;

}

// The main function to convert infix expression

// to postfix expression

void infixToPostfix(string s)

{

    stack<char> st;

    string result;

    for (int i = 0; i < s.length(); i++) {

        char c = s[i];

        // If the scanned character is

        // an operand, add it to output string.

        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')

            || (c >= '0' && c <= '9'))

            result += c;

        // If the scanned character is an

        // ‘(‘, push it to the stack.

        else if (c == '(')

            st.push('(');

        // If the scanned character is an ‘)’,

        // pop and add to output string from the stack

        // until an ‘(‘ is encountered.

        else if (c == ')') {

            while (st.top() != '(') {

                result += st.top();

                st.pop();

            }

            st.pop();

        }

        // If an operator is scanned

        else {

            while (!st.empty()

                   && prec(s[i]) <= prec(st.top())) {

                result += st.top();

                st.pop();

            }

            st.push(c);

        }

    }

    // Pop all the remaining elements from the stack

    while (!st.empty()) {

        result += st.top();

        st.pop();

    }

    cout << result << endl;

}

// Driver code

int main()

{

    string exp = "a+b*(c^d-e)^(f+g*h)-i";

    // Function call

    infixToPostfix(exp);

    return 0;

}