Problem: For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique.

* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print the value with the specified $key$. Print 0 if there is no such element.
* delete($key$): Delete the element with the specified $key$.

Constraints

* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consits of lower case letters

Input

The input is given in the following format.


$q$
$query_1$
$query_2$
:
$query_q$


Each query $query_i$ is given by


0 $key$ $x$


or


1 $key$


or


2 $key$


where the first digits 0, 1 and 2 represent insert, get and delete operations respectively.

Output

For each get operation, print an integer in a line.

Example

Input

8
0 blue 4
0 red 1
0 white 5
1 red
1 blue
2 red
1 black
1 red


Output

1
4
0
0

Solution (in CPP):

Solution:
#include <iostream>
#include <iterator>
#include <map>
#include <string>

using Key = std::string;

int main(void) {
    int q;
    std::cin >> q;

    std::map<Key, int> my_map{};
    for (int i = 0; i < q; ++i) {
        int kind;
        std::cin >> kind;

        Key key;
        switch (kind) {
            case 0:
                /**
                 * insert
                 * add key-x to my_map
                 * if exist, update it
                 */
                int x;
                std::cin >> key >> x;
                if (my_map.find(key) == my_map.end()) {
                    my_map.insert(std::make_pair(key, x));
                } else {
                    my_map[key] = x;
                }
                break;
            case 1:
                /**
                 * get
                 * output value whose key is it
                 * but no-exist case, output 0
                 */
                std::cin >> key;
                if (my_map.find(key) != my_map.end()) {
                    std::cout << my_map[key] << std::endl;
                } else {
                    std::cout << 0 << std::endl;
                }
                break;
            case 2:
                /**
                 * delete
                 * delete member whose key is it
                 */
                std::cin >> key;
                my_map.erase(key);
                break;
        }
    }
}