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 <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007
int U[100001];
set<int> R[100001];

int find(int x) {
  if (U[x] == x) return x;
  return U[x] = find(U[x]);
}

void unite(int x, int y) {
  x = find(x), y = find(y);
  if (x == y) return;
  if (R[x].size() < R[y].size()) swap(x, y);
  U[y] = x;
  for (int e : R[y]) R[x].insert(e);
  R[y].clear();
}

bool same(int x, int y) {
  return find(x) == find(y);
}
inline int mins(int v) {
  return (26-(v%26))%26;
}

int N, M;
string S;
int A[100001];
bool done[100001];
void add(int p, int v) {
  v = (26+v)%26;
  A[p] = (A[p]+v)%26;
}
bool go(int pos, int v) {
  v = (26+v)%26;
  if (R[find(pos)].empty()) return false;
  int a = A[pos];
  int left = *R[find(pos)].begin(), right = *R[find(pos)].rbegin();
  if (!done[left]) {
    add(left, a-v);
    add(pos, v-a);
    assert(A[pos] == v);
    return true;
  }
  if (!done[right]) {
    add(right, a-v);
    add(pos, v-a);
    assert(A[pos] == v);
    return true;
  }
  return false;
}

signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> S >> M;
  N = S.length();
  rep(i, N+1) U[i] = i, R[i].insert(i);
  rep(i, M) {
    int l, r;
    cin >> l >> r;
    l--;
    unite(l, r);
  }
  rep(i, N) A[i] = S[i]-(i>0?S[i-1]:'a');
  rep(i, N) A[i] = ((A[i]%26)+26)%26;
  A[N] = 0;
  N++;
  for (int p=N/2-1; p>=0; p--) {
    int right = N-1-p;
    unite(p, right);
  }
  if (N % 2 == 1) {
    int pos = (N-1)/2;
    // pos -> 0
    done[pos] = true;
    if (A[pos] != 0) {
      if (!go(pos, 0)) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  int hoge = N/2;
  for (int p=N/2-1; p>=1; p--) {
    int right = N-1-p;
    done[p] = done[right] = true;
    if (A[p] == mins(A[right])) continue;
    if (go(p, mins(A[right])) || go(right, mins(A[p]))) {
    }
    else {
      cout << "NO\n";
      return 0;
    }
  }
  cout << "YES\n";
  return 0;
}