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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i, n) for (int i = 0; i < (int) (n); i++)
#define reps(i, n) for (int i = 1; i <= (int) (n); i++)
#define all(x) (x).begin(), (x).end()
#define uniq(x) (x).erase(unique(all(x)), (x).end())
#define bit(n) (1LL << (n))
#define cdiv(a, b) (((a) - 1) / (b) + 1)
#define dump(x) cerr << #x " = " << (x) << endl
using vint = vector<int>;
using vvint = vector<vint>;
using pint = pair<int, int>;
using vpint = vector<pint>;
template<typename T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
constexpr double PI = 3.1415926535897932384626433832795028;
constexpr int DY[9] = {0, 1, 0, -1, 1, 1, -1, -1, 0};
constexpr int DX[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
int gcd(int a, int b) {
    while (b) { swap(a %= b, b); }
    return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
int sign(int x) { return (x > 0) - (x < 0); }
template<typename T> void fin(T mes) {
    cout << mes << endl;
    exit(0);
}
template<typename T, typename U> bool chmax(T &a, const U &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> bool chmin(T &a, const U &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &rhs) {
    os << "(" << rhs.first << ", " << rhs.second << ")";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const vector<T> &rhs) {
    os << "{";
    for (auto itr = rhs.begin(); itr != rhs.end(); itr++) {
        os << *itr << (next(itr) != rhs.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const deque<T> &rhs) {
    os << "{";
    for (auto itr = rhs.begin(); itr != rhs.end(); itr++) {
        os << *itr << (next(itr) != rhs.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const set<T> &rhs) {
    os << "{";
    for (auto itr = rhs.begin(); itr != rhs.end(); itr++) {
        os << *itr << (next(itr) != rhs.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T, typename U> ostream &operator<<(ostream &os, const map<T, U> &rhs) {
    os << "{";
    for (auto itr = rhs.begin(); itr != rhs.end(); itr++) {
        os << *itr << (next(itr) != rhs.end() ? ", " : "");
    }
    os << "}";
    return os;
}
struct setup {
    static constexpr int PREC = 20;
    setup() {
        cout << fixed << setprecision(PREC);
        cerr << fixed << setprecision(PREC);
    };
} setup;

template<int MOD = 1000000007>
struct mod_int {
    int val;
    mod_int(long long val_ = 0) : val(val_ % MOD) { if (val < 0) { val += MOD; }}
    bool operator==(const mod_int &rhs) const { return val == rhs.val; }
    bool operator!=(const mod_int &rhs) const { return std::rel_ops::operator!=(*this, rhs); }
    mod_int &operator+=(const mod_int &rhs) {
        if ((val += rhs.val) >= MOD) { val -= MOD; }
        return *this;
    }
    mod_int &operator-=(const mod_int &rhs) {
        if ((val += MOD - rhs.val) >= MOD) { val -= MOD; }
        return *this;
    }
    mod_int &operator*=(const mod_int &rhs) {
        val = (int) ((long long) val * rhs.val % MOD);
        return *this;
    }
    mod_int &operator/=(const mod_int &rhs) { return *this *= rhs.inv(); }
    mod_int operator+() const { return *this; }
    mod_int operator-() const { return -val; }
    mod_int operator++() { return *this += 1; }
    mod_int operator--() { return *this -= 1; }
    mod_int operator++(signed) {
        const mod_int ret(*this);
        ++*this;
        return ret;
    }
    mod_int operator--(signed) {
        const mod_int ret(*this);
        --*this;
        return ret;
    }
    mod_int operator+(const mod_int &rhs) const { return mod_int(*this) += rhs; }
    mod_int operator-(const mod_int &rhs) const { return mod_int(*this) -= rhs; }
    mod_int operator*(const mod_int &rhs) const { return mod_int(*this) *= rhs; }
    mod_int operator/(const mod_int &rhs) const { return mod_int(*this) /= rhs; }
    mod_int inv() const {
        assert(val != 0);
        int a = val, b = MOD, x = 1, u = 0;
        while (b) {
            int t = a / b;
            std::swap(a -= t * b, b);
            std::swap(x -= t * u, u);
        }
        return x;
    }
    mod_int pow(long long n) const {
        if (n < 0) { return pow(-n).inv(); }
        mod_int ret = 1, mul = *this;
        while (n) {
            if (n & 1) { ret *= mul; }
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }
    friend std::istream &operator>>(std::istream &is, mod_int &rhs) {
        long long v;
        is >> v;
        rhs = v;
        return is;
    }
    friend std::ostream &operator<<(std::ostream &os, const mod_int &rhs) { return os << rhs.val; }
    struct combination {
        std::vector<mod_int> fact{1, 1}, f_inv{1, 1}, inv{0, 1};
        void calc(int n) {
            while (fact.size() <= n) {
                int i = fact.size();
                fact.push_back(fact[i - 1] * i);
                inv.push_back(-inv[MOD % i] * (MOD / i));
                f_inv.push_back(f_inv[i - 1] * inv[i]);
            }
        }
        mod_int P(int n, int r) { return r < 0 || n < r ? 0 : (calc(n), fact[n] * f_inv[n - r]); }
        mod_int C(int n, int r) { return r < 0 || n < r ? 0 : (calc(n), fact[n] * f_inv[r] * f_inv[n - r]); }
        mod_int H(int n, int r) { return C(n + r - 1, r); }
    };
};
using mint = mod_int<26>;

template<typename M>
struct fenwick_tree {
    using T = typename M::T;
    int n;
    std::vector<T> data;
    fenwick_tree(int n) : n(n), data(n + 1, M::id()) {}
    void add(int i, const T &x) { for (i++; i <= n; i += i & -i) { data[i] = M::op(data[i], x); }}
    T get_sum(int i) const {
        T ret = M::id();
        for (; i > 0; i -= i & -i) { ret = M::op(ret, data[i]); }
        return ret;
    }
    T get_sum(int l, int r) const {
        return M::op(get_sum(r), M::inv(get_sum(l)));
    }
};

template<typename M>
struct fenwick_tree_range {
    using T = typename M::T;
    fenwick_tree<M> ft;
    fenwick_tree_range(int n) : ft(n + 1) {}
    void add(int l, int r, const T &x) { ft.add(l, x), ft.add(r, -x); }
    T operator[](int i) const { return ft.get_sum(i + 1); }
};

struct rsq {
    using T = mint;
    static T id() { return 0; }
    static T op(const T &a, const T &b) { return a + b; }
    static T inv(const T &a) { return -a; }
};

string S;
int N;
int L[110000], R[110000];
priority_queue_rev<int> rs[110000];
signed main() {
    cin >> S >> N;
    int K = S.length();
    rep(i, N) {
        cin >> L[i] >> R[i];
        if (L[i] - 1 < K - R[i]) { rs[L[i] - 1].push(min(R[i], K - R[i])); }
        else if (L[i] - 1 > K - R[i]) { rs[K - R[i]].push(min(L[i] - 1, K - L[i] + 1)); }
    }
    fenwick_tree_range<rsq> ft(K / 2);
    rep(i, K / 2) { ft.add(i, i + 1, S[i] - S[K - i - 1]); }
    rep(i, K / 2) {
        if (rs[i].empty()) {
            if (ft[i] != 0) { fin("NO"); }
            continue;
        }
        int cur = rs[i].top();
        rs[i].pop();
        ft.add(i, cur, -ft[i]);
        while (!rs[i].empty()) {
            if (cur < rs[i].top()) { rs[cur].push(rs[i].top()); }
            cur = rs[i].top(), rs[i].pop();
        }
    }
    cout << "YES" << endl;
}