Problem: 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.

Examples

Input

bixzja
2
2 3
3 6


Output

YES


Input

abc
1
2 2


Output

NO


Input

cassert
4
1 2
3 4
1 1
2 2


Output

YES

Solution (in CPP):

Solution:
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);

ll mod_pow(ll x, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * x % mod;
		x = x * x % mod; n >>= 1;
	}
	return res;
}
struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n % mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a * a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 20;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}

struct uf {
private:
	vector<int> par, ran;
public:
	uf(int n) {
		par.resize(n, 0);
		ran.resize(n, 0);
		rep(i, n) {
			par[i] = i;
		}
	}
	int find(int x) {
		if (par[x] == x)return x;
		else return par[x] = find(par[x]);
	}
	void unite(int x, int y) {
		x = find(x), y = find(y);
		if (x == y)return;
		if (ran[x] < ran[y]) {
			par[x] = y;
		}
		else {
			par[y] = x;
			if (ran[x] == ran[y])ran[x]++;
		}
	}
	bool same(int x, int y) {
		return find(x) == find(y);
	}
};

void solve() {
	string s; cin >> s;
	int n; cin >> n;
	int m = s.size();
	vector<P> v(n);
	rep(i, n) {
		cin >> v[i].first >> v[i].second;
		v[i].first--; v[i].second--;
	}
	if (m == 1) {
		cout << "YES\n"; return;
	}
	vector<int> a(m / 2);
	rep(i, m / 2) {
		int dif = s[m - 1 - i] - s[i];
		if (dif < 0)dif += 26;
		a[i] = dif;
	}
	vector<int>d;
	d.push_back(a[0]);
	rep(i, m / 2 - 1) {
		int dif = a[i + 1] - a[i]; if (dif < 0)dif += 26;
		d.push_back(dif);
	}
	d.push_back(26 - a.back());
	uf u(d.size());
	rep(i, n) {
		int le = v[i].first, ri = v[i].second;
		if (le >= m / 2) {
			le = m - 1 - le;
			ri = m - 1 - ri;
			swap(le, ri);
		}
		if (ri >= m / 2) {
			int cl = m / 2 - le;
			int cr = ri - (m - 1) / 2;
			le = m / 2 - max(cl, cr);
			ri = m / 2 - min(cl, cr)-1;
		}
		if (ri == m / 2 && m % 2)ri--;
		u.unite(le, ri + 1);
	}
	vector<vector<int>> ids(d.size());
	rep(i, d.size()) {
		ids[u.find(i)].push_back(i);
	}
	for (vector<int> c : ids) {
		int sum = 0;
		for (int id : c) {
			sum += d[id];
		}
		if (sum % 26) {
			cout << "NO\n"; return;
		}
	}
	cout << "YES\n";
}




signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//init();
	//int t; cin >> t; rep(i, t)
	solve();
	return 0;
}