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

using ll=int64_t;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#define REACH cerr<<"reached line "<<__LINE__<<endl
#define DMP(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl
#define ZERO(x) memset(x,0,sizeof(x))

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"[";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"]";
	return os;
}

int read(){
	int i;
	scanf("%" SCNd64,&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(int x,int suc=1){
	printf("%" PRId64,x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341919];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341919];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(a>b)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

const int inf=LLONG_MAX/3;

const int Nmax=114514;
int val[Nmax];
vi g[Nmax];
bool vis[Nmax];
int dfs(int v){
	if(vis[v])return 0;
	vis[v]=true;
	int res=val[v];
	for(auto to:g[v])
		res+=dfs(to);
	return res;
}

void No(){
	cout<<"NO"<<endl;
	exit(0);
}
void Yes(){
	cout<<"YES"<<endl;
	exit(0);
}

signed main(){
	string s=readString();
	int n=s.size();
	REP(i,n/2)
		val[i]=s[i]-s[n-1-i];
	for(int i=n/2;i>=1;i--)
		val[i]=val[i]-val[i-1];
	int q=read();
	REP(i,q){
		int l=read()-1,r=read();
		l=min(l,n-l);
		r=min(r,n-r);
		if(l>r)swap(l,r);
		g[l].PB(r);
		g[r].PB(l);
	}
	REP(i,n/2+1)if(!vis[i])
		if(dfs(i)%26)
			No();
	Yes();
}