Problem:
Score : 1000 points

Problem Statement

Ringo has a string S.

He can perform the following N kinds of operations any number of times in any order.

Operation i: For each of the characters from the L_i-th through the R_i-th characters in S, replace it with its succeeding letter in the English alphabet. (That is, replace a with b, replace b with c and so on.) For z, we assume that its succeeding letter is a.

Ringo loves palindromes and wants to turn S into a palindrome.
Determine whether this is possible.

Constraints

1 \leq |S| \leq 10^5

S consists of lowercase English letters.

1 \leq N \leq 10^5

1 \leq L_i \leq R_i \leq |S|

Input

Input is given from Standard Input in the following format:

S
N
L_1 R_1
L_2 R_2
:
L_N R_N

Output

Print YES if it is possible to turn S into a palindrome; print NO if it is impossible.

Sample Input 1

bixzja
2
2 3
3 6

Sample Output 1

YES

For example, if we perform Operation 1, 2 and 1 in this order, S changes as bixzja → bjyzja → bjzakb → bkaakb and becomes a palindrome.

Sample Input 2

abc
1
2 2

Sample Output 2

NO

Sample Input 3

cassert
4
1 2
3 4
1 1
2 2

Sample Output 3

YES

Solution in Rust (1.15.1) (judge verdict: Accepted)

//---------- begin union_find ----------
mod union_find {
    pub struct UF {
        p: Vec<i32>,
    }
    #[allow(dead_code)]
    impl UF {
        pub fn new(n: usize) -> UF {
            UF {p: vec![-1; n] }
        }
        pub fn init(&mut self) {
            for p in self.p.iter_mut() {
                *p = -1;
            }
        }
        pub fn root(&mut self, mut x: usize) -> usize {
            while self.p[x] >= 0 {
                x = self.p[x] as usize;
            }
            x
        }
        pub fn same(&mut self, x: usize, y: usize) -> bool {
            self.root(x) == self.root(y)
        }
        pub fn unite(&mut self, mut x: usize, mut y: usize) {
            x = self.root(x);
            y = self.root(y);
            if x == y {
                return;
            }
            if self.p[x] > self.p[y] {
                let s = x;
                x = y;
                y = s;
            }
            self.p[x] += self.p[y];
            self.p[y] = x as i32;
        }
        pub fn get_size(&mut self, x: usize) -> usize {
            let r = self.root(x);
            (-self.p[r]) as usize
        }
    }
}
//---------- end union_find ----------
//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

fn run() {
    input! {
        s: String,
        m: usize,
        p: [(usize1, usize); m],
    }
    let a: Vec<u32> = s.chars().map(|c| c as u32 - 'a' as u32).collect();
    let n = a.len();
    let mut u = union_find::UF::new(n + 1);
    for (l, r) in p {
        u.unite(l, r);
    }
    for i in 0..n {
        u.unite(i, n - i);
    }
    let mut dp = vec![0; n + 1];
    dp[u.root(0)] += a[0];
    for i in 1..n {
        dp[u.root(i)] += (a[i] - a[i - 1] + 26) % 26;
    }
    dp[u.root(n)] += 26 - a[n - 1];
    for i in 0..(n + 1) {
        if dp[i] % 26 != 0 {
            println!("NO");
            return;
        }
    }
    println!("YES");
}

fn main() {
    run();
}