Problem: You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.

Examples

Input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2


Output

2
6
-1
4

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.io.IOException;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author lebegio
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        StreamTokenizer in = new StreamTokenizer(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public StreamTokenizer st;

        public void solve(int testNumber, StreamTokenizer st, PrintWriter out) {
            this.st = st;
            int n = nextInt();
            int m = nextInt();

            long[] arr = new long[n];
            int[] pos = new int[n];
            for (int i = 0; i < n; i++) {
                long p = nextLong();
                arr[i] = p;
                if (i == 0) {
                    pos[i] = 0;
                } else {
                    if (arr[i - 1] == p) {
                        pos[i] = pos[i - 1];
                    } else {
                        pos[i] = i;
                    }
                }
            }
            for (int i = 0; i < m; i++) {
                int l = nextInt() - 1;
                int r = nextInt() - 1;
                long x = nextLong();
                if (l == r) {
                    if (arr[r] != x) out.println(r + 1);
                    else out.println(-1);
                    continue;
                }


                if (arr[r] != x) out.println(r + 1);
                else {
                    if (pos[r] == pos[l]) out.println(-1);
                    else {
                        int inx = pos[r];
                        if (l < inx) {
                            out.println(inx);
                        } else {
                            out.println(-1);
                        }
                    }
                }


            }
        }

        int nextInt() {
            try {
                st.nextToken();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return (int) st.nval;
        }

        long nextLong() {
            try {
                st.nextToken();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return (long) st.nval;
        }

    }
}