Problem:
Score : 300 points

Problem Statement

E869120 found a chest which is likely to contain treasure.

However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.

He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with ?.

One more thing he found is a sheet of paper with the following facts written on it:

Condition 1: The string S contains a string T as a contiguous substring.

Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.

Print the string S.

If such a string does not exist, print UNRESTORABLE.

Constraints

1 \leq |S'|, |T| \leq 50

S' consists of lowercase English letters and ?.

T consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T'

Output

Print the string S.

If such a string does not exist, print UNRESTORABLE instead.

Sample Input 1

?tc????
coder

Sample Output 1

atcoder

There are 26 strings that satisfy Condition 1: atcoder, btcoder, ctcoder,..., ztcoder.
Among them, the lexicographically smallest is atcoder, so we can say S = atcoder.

Sample Input 2

??p??d??
abc

Sample Output 2

UNRESTORABLE

There is no string that satisfies Condition 1, so the string S does not exist.

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::BTreeSet;
use std::collections::HashSet;
use std::cmp;
use std::f64::consts;
use std::cmp::Ordering;

fn main() {
    let input: String = read();
    let s_vec: Vec<char> = input.chars().collect();
    let subs: String = read();
    let subs_vec: Vec<char> = subs.chars().collect();

    let first_subs = subs_vec[0];

    let mut yes = false;

    let mut ans_vec: Vec<char> = s_vec.clone();

    let mut answer = "".to_string();

    for (i, &s) in s_vec.iter().enumerate() {
        if s == first_subs || s == '?' {
            let mut count: usize = 0;
            for j in i..s_vec.len() {
                let subs_index = j-i;
                if subs_index > subs_vec.len() -1 {
                    break;
                }
                if s_vec[j] != subs_vec[subs_index] && s_vec[j] != '?' {
                    break;
                } else {
                    count+=1;
                    ans_vec[j] = subs_vec[subs_index].clone();
                }
            }

            if count == subs_vec.len() {
                let mut tmpanswer = "".to_string();
                for &s in ans_vec.iter() {
                    if s == '?' {
                        tmpanswer = format!("{}{}", tmpanswer, 'a'.to_string());
                    } else {
                        tmpanswer = format!("{}{}", tmpanswer, s.to_string());
                    }
                }
                println!("{}", tmpanswer);
                if tmpanswer < answer {
                    answer = tmpanswer;
                }
                yes = true;
                break;
            } else {
                ans_vec = s_vec.clone();
            }
        }
    }

    if yes {
        println!("{}", answer);
    } else {
        println!("UNRESTORABLE");
    }
}

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

// 最大公約数
fn gcd(a: i32, b: i32) -> i32 {
    match b {
        0 => a,
        _ => gcd(b, a % b)
    }
}

// 最小公倍数
fn lcm(a: i32, b: i32) -> i32 {
    a * b / gcd(a, b)
}

// 階乗
fn kaijou(n: i32)->i32 {
    if n == 1 {
        return n;
    }
    return n * kaijou(n-1);
}

// 順列全列挙
pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    /// let mut data = [1, 2, 3];
    //    let mut permutations = Vec::new();
    //
    //    loop {
    //        permutations.push(data.to_vec());
    //        if !data.next_permutation() {
    //            break;
    //        }
    //    }
    //
    //    println!("{:?}", permutations);
    //    // [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }
}

//iより小さい数字の数 = lower_bound(&a, i) as i64;
//iより大きい数字の数 = n - upper_bound(&c, i) as i64;
fn lower_bound<T: Ord>(vec: &Vec<T>, x: T) -> usize {
    let (mut left, mut right): (i64, i64) = (-1, vec.len() as i64);

    while (right - left) > 1 {
        let mid = (right + left) / 2;

        if x <= vec[mid as usize] {
            right = mid;
        } else {
            left = mid;
        }
    }

    return right as usize;
}

fn upper_bound<T: Ord>(vec: &Vec<T>, x: T) -> usize {
    let (mut left, mut right): (i64, i64) = (-1, vec.len() as i64);

    while (right - left) > 1 {
        let mid = (right + left) / 2;

        if x < vec[mid as usize] {
            right = mid;
        } else {
            left = mid;
        }
    }

    return right as usize;
}