Problem: In this problem, we will deal with binary strings. Each character of a binary string is either a 0 or a 1. We will also deal with substrings; recall that a substring is a contiguous subsequence of a string. We denote the substring of string s starting from the l-th character and ending with the r-th character as s[l ... r]. The characters of each string are numbered from 1.

We can perform several operations on the strings we consider. Each operation is to choose a substring of our string and replace it with another string. There are two possible types of operations: replace 011 with 110, or replace 110 with 011. For example, if we apply exactly one operation to the string 110011110, it can be transformed into 011011110, 110110110, or 110011011.

Binary string a is considered reachable from binary string b if there exists a sequence s_1, s_2, ..., s_k such that s_1 = a, s_k = b, and for every i ∈ [1, k - 1], s_i can be transformed into s_{i + 1} using exactly one operation. Note that k can be equal to 1, i. e., every string is reachable from itself.

You are given a string t and q queries to it. Each query consists of three integers l_1, l_2 and len. To answer each query, you have to determine whether t[l_1 ... l_1 + len - 1] is reachable from t[l_2 ... l_2 + len - 1].

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of string t.

The second line contains one string t (|t| = n). Each character of t is either 0 or 1.

The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Then q lines follow, each line represents a query. The i-th line contains three integers l_1, l_2 and len (1 ≤ l_1, l_2 ≤ |t|, 1 ≤ len ≤ |t| - max(l_1, l_2) + 1) for the i-th query.

Output

For each query, print either YES if t[l_1 ... l_1 + len - 1] is reachable from t[l_2 ... l_2 + len - 1], or NO otherwise. You may print each letter in any register.

Example

Input


5
11011
3
1 3 3
1 4 2
1 2 3


Output


Yes
Yes
No

Solution (in JAVA):

Solution:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.Reader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class C1320D {

    static long modulo1 = (long) 1e9 + 9;
    static long modulo2 = (long) 1e9 + 7;
    static long r1 = 3;
    static long r2 = 5;
    static int[] bits1 = new int[]{1, 2};
    static int[] bits2 = new int[]{2, 3};

    public static void main(String[] args) {
        var scanner = new BufferedScanner();
        var writer = new PrintWriter(new BufferedOutputStream(System.out));

        var n = scanner.nextInt();
        var t = scanner.next();
        var count0 = new int[n + 1];
        {
            count0[0] = 0;
            for (int len = 1; len <= n; len++) {
                count0[len] = count0[len - 1] + (t.charAt(len - 1) == '0' ? 1 : 0);
            }
        }
        var power1 = new long[n + 1];
        var power2 = new long[n + 1];
        power1[0] = 1;
        power2[0] = 1;
        for (int i = 1; i <= n; i++) {
            power1[i] = power1[i - 1] * r1 % modulo1;
            power2[i] = power2[i - 1] * r2 % modulo2;
        }
        var hashEven1 = new long[n + 1][];
        var hashOdd1 = new long[n + 1][];
        calcHash(t, n, power1, hashEven1, hashOdd1, count0, modulo1, r1, bits1);
        var hashEven2 = new long[n + 1][];
        var hashOdd2 = new long[n + 1][];
        calcHash(t, n, power2, hashEven2, hashOdd2, count0, modulo2, r2, bits2);
        var q = scanner.nextInt();
        for (int qi = 0; qi < q; qi++) {
            var l1 = scanner.nextInt() - 1;
            var l2 = scanner.nextInt() - 1;
            var len = scanner.nextInt();
            var ans = false;
            // 首先，0的个数得一样多
            if (count0[l1 + len] - count0[l1] == count0[l2 + len] - count0[l2]) {
                // 其次，从各子串开头算起的0的下标得对应位同奇偶。
                var fp11 = fingerprint(l1, l1 + len - 1, power1, hashEven1, hashOdd1, count0, modulo1);
                var fp21 = fingerprint(l2, l2 + len - 1, power1, hashEven1, hashOdd1, count0, modulo1);
                var fp12 = fingerprint(l1, l1 + len - 1, power2, hashEven2, hashOdd2, count0, modulo2);
                var fp22 = fingerprint(l2, l2 + len - 1, power2, hashEven2, hashOdd2, count0, modulo2);
//                System.out.println(String.format("[%d] %d,%d %d,%d", qi + 1, fp11, fp21, fp12, fp22));
                if (fp11 == fp21 && fp12 == fp22) {
                    ans = true;
                }
            }
            writer.println(ans ? "Yes" : "No");
        }

        scanner.close();
        writer.flush();
        writer.close();
    }

    private static void calcHash(String t, int n, long[] power, long[][] hashEven, long[][] hashOdd, int[] count0,
            long modulo, long r, int[] bits) {
        var limit = r * modulo;
//        System.out.println(String.format("modulo=%d,r=%d,bits=%s", modulo, r, Arrays.toString(bits)));
        for (int len = 1; len <= n; len <<= 1) {
            hashEven[len] = new long[n];
            hashOdd[len] = new long[n];
            var he = 0L;
            var ho = 0L;
            for (int i = 0; i < n; i++) {
                if (t.charAt(i) == '0') {
                    var bitEven = bits[i & 1]; // 对偶数开头子字符串来说，i是偶数在子字符串也是偶数位，用0表示；否则用1。
                    var bitOdd = bits[1 - (i & 1)]; // 对奇数位开头的子字符串来说，i是偶数在子字符串是奇数位，用1表示；否则用0。
                    he = (he * r + bitEven) % modulo;
                    ho = (ho * r + bitOdd) % modulo;
                }
                if (i >= len && t.charAt(i - len) == '0') {
                    var bitEven0 = bits[(i - len) & 1];
                    var bitOdd0 = bits[1 - ((i - len) & 1)];
                    var cnt0 = count0[i + 1] - count0[i + 1 - len];
                    he = (he + limit - power[cnt0] * bitEven0) % modulo; // power[cnt0]*bitEven0<modulo*r
                    ho = (ho + limit - power[cnt0] * bitOdd0) % modulo; // power[cnt0]*bitOdd0<modulo*r
                }
                if (i >= len - 1) {
                    hashEven[len][i - len + 1] = he;
                    hashOdd[len][i - len + 1] = ho;
                }
                if (he < 0 || ho < 0) {
                    throw new RuntimeException("wtf");
                }
            }
        }
    }

    private static long fingerprint(int start, int end, long[] power, long[][] hashEven, long[][] hashOdd, int[] count0,
            long modulo) {
        if ((start & 1) == 0) {
            return fingerprint(start, end, power, hashEven, count0, modulo);
        } else {
            return fingerprint(start, end, power, hashOdd, count0, modulo);
        }
    }

    private static long fingerprint(int start, int end, long[] power, long[][] hash, int[] count0, long modulo) {
        var len = 1 << (int) (Math.log(end - start + 1) / Math.log(2));
        var fp = 0L;
        while (start <= end) {
            var cnt0 = count0[start + len] - count0[start];
            fp = (fp * power[cnt0] + hash[len][start]) % modulo;
            if (fp < 0) {
                throw new RuntimeException("wtf");
            }
            start += len;
            while (len > 0 && start + len - 1 > end) {
                len >>= 1;
            }
        }
        return fp;
    }

    public static class BufferedScanner {
        BufferedReader br;
        StringTokenizer st;

        public BufferedScanner(Reader reader) {
            br = new BufferedReader(reader);
        }

        public BufferedScanner() {
            this(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

    }

    static long gcd(long a, long b) {
        if (a < b) {
            return gcd(b, a);
        }
        while (b > 0) {
            long tmp = b;
            b = a % b;
            a = tmp;
        }
        return a;
    }

    static long inverse(long a, long m) {
        long[] ans = extgcd(a, m);
        return ans[0] == 1 ? (ans[1] + m) % m : -1;
    }

    private static long[] extgcd(long a, long m) {
        if (m == 0) {
            return new long[]{a, 1, 0};
        } else {
            long[] ans = extgcd(m, a % m);
            long tmp = ans[1];
            ans[1] = ans[2];
            ans[2] = tmp;
            ans[2] -= ans[1] * (a / m);
            return ans;
        }
    }

    private static List<Integer> primes(double upperBound) {
        var limit = (int) Math.sqrt(upperBound);
        var isComposite = new boolean[limit + 1];
        var primes = new ArrayList<Integer>();
        for (int i = 2; i <= limit; i++) {
            if (isComposite[i]) {
                continue;
            }
            primes.add(i);
            int j = i + i;
            while (j <= limit) {
                isComposite[j] = true;
                j += i;
            }
        }
        return primes;
    }


}