Codeforce Round #215 Div2 C

C. Sereja and Algorithm

time limit per test 1 second

memory limit per test 256 megabytes

这道题没出来真的是自己,啊改了一点就没管了，结果xx也可以的哎- -，当时也用种情况交过，去掉特判后又忘记加t<=2了，B！

Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q₁q₂... q_k. The algorithm consists of two steps:

1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2.
2. Rearrange the letters of the found subsequence randomly and go to step 1.

 

Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string.

Sereja wants to test his algorithm. For that, he has string s = s₁s₂... s_n, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring s_lis_li + 1... s_ri (1 ≤ l_i ≤ r_i ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (l_i, r_i) determine if the algorithm works correctly on this test or not.

Input

The first line contains non-empty string s, its length (n) doesn't exceed 10⁵. It is guaranteed that string s only contains characters: 'x', 'y', 'z'.

The second line contains integer m (1 ≤ m ≤ 10⁵) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.

Sample test(s)

Input

zyxxxxxxyyz 5 5 5 1 3 1 11 1 4 3 6

Output

YES YES NO YES NO

Note

In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <queue>

using namespace std;
#define INF 0x7fffffff
#define maxm 1001
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); i++)
#define re return
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define sqrt(x) sqrt(abs(x))
#define y0 y3487465
//#define y1 y8687969
//#define fill(x,y) memset(x,y,sizeof(x))

typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef double D;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef vector<vi> vvi;

template<class T> T abs(T x) { re x > 0 ? x : -x; }

const int maxn = 110005;

int n, m, t, k;
int vis[maxn];
int x[maxn], y[maxn], z[maxn];
char s[maxn];
int main(){
    scanf("%s", s);
    n = strlen(s);
    for (int i = 1; i <= n; i++){
        x[i] = x[i - 1]; y[i] = y[i - 1]; z[i] = z[i - 1];
        if (s[i-1] == 'x')x[i]++;
        if (s[i-1] == 'y')y[i]++;
        if (s[i-1] == 'z')z[i]++;
    }
    scanf("%d", &m);
    while (m--){
        int a, b;
        scanf("%d%d", &a, &b);
        t = b - a + 1;
        int t1=x[b]-x[a-1], t2=y[b]-y[a-1], t3=z[b]-z[a-1];
        int m1, m2, m3,flag=1;
        m1 = max(max(t1, t2), t3);
        m3 = min(min(t1, t2), t3);
        m2 = t1 + t2 + t3 - m1 - m3;
        if ((t - 1) / 3 + 1 < m1)flag = 0;
        if ((t - 1-1) / 3 + 1 < m2)flag = 0;
        if ((t - 1-1-1) / 3 + 1 < m3)flag = 0;
        if (flag||t<=2)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}