Vladik had started reading a complicated book about algorithms containing n pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation P = [p1, p2, ..., pn], where pi denotes the number of page that should be read i-th in turn.

Sometimes Vladik’s mom sorted some subsegment of permutation P from position l to position r inclusive, because she loves the order. For every of such sorting Vladik knows number x — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has px changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.

Input

First line contains two space-separated integers n, m (1 ≤ n, m ≤ 104) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.

Second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ n) — permutation P. Note that elements in permutation are distinct.

Each of the next m lines contains three space-separated integers li, ri, xi (1 ≤ li ≤ xi ≤ ri ≤ n) — left and right borders of sorted subsegment in i-th sorting and position that is interesting to Vladik.

Output

For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.

Examples

Input

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


Output

Yes
No
Yes
Yes
No


Input

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


Output

Yes
No
Yes
No
Yes

Note

Explanation of first test case: 

  1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 
  2. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No". 
  3. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes". 
  4. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes". 
  5. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No". 
import java.io.*;
import java.math.*;
import java.security.KeyStore.Entry;
import java.util.*;


 public class Main {
	private InputStream is;
	private PrintWriter out;
	int time = 0, dp[][], start[], red[], black[], MOD = (int)(1e9+7), arr[];
	int MAX = 200000, N, K;
	ArrayList<Integer>[] amp;
	boolean b[];
	Pair prr[];
	HashSet<Integer> hs = new HashSet<>();
	void solve() {
		int n = ni(), m = ni();
		int arr[] = na(n);
		while(m-->0){
			int l = ni()-1, r = ni()-1, x = ni()-1;
			int cnt = 0;
			for(int i = l;i<=r;i++){
				if(arr[i]<arr[x]) cnt++;
			}
			if(l+cnt==x){
				out.println("Yes");
			}
			else out.println("No");
		}
	}
	void recur(int n, int k){
		if(k<0) return;
		hs.add(K-k);
		if(n>=0 && dp[n][k]!=-1) return;
		if(n>=0) dp[n][k] = 1;
		for(int i = n+1;i<N;i++){
			recur(i,k-arr[i]);
		}
	}
	long modInverse(long a, long mOD2){
        return  power(a, mOD2-2, mOD2);
	}
	long power(long x, long y, long m)
	{
		if (y == 0)
    return 1;
	long p = power(x, y/2, m) % m;
	p = (p * p) % m;

	return (y%2 == 0)? p : (x * p) % m;
	}
	boolean isPrime(int x){
		for(int i = 2;i*1L*i<=x;i++) if(x%i==0) return false;
		return true;
	}
	public String isPossible(int b[]){
		String ans = "Possible";
		int n = b.length;
		Arrays.sort(b);
		if(n==1) return "Possible";
		if(b[0]==1) return "Impossible";
		if(b[0]>=n) return ans;
		int lcm = 1;
		for(int i : b){
			long gcd = gcd(lcm,i);
			lcm *= (i/gcd);
		}
		int cnt = 0;
		for(int i : b) cnt+=lcm/i;
		if(cnt>lcm) return "Impossible";
		return ans;
	}
	public long gcd(long a, long b){
		if(b==0) return a;
		return gcd(b,a%b);
	}
	void failFn(String str, int arr[]){
		int len = 0;
		arr[0] = 0;
		int i = 1;
		while(i<arr.length){
			if(str.charAt(i)==str.charAt(len)){
				arr[i++] = ++len;
				continue;
			}
			if(len == 0){
				arr[i] = len;
				i++;
				continue;
			}
			if(str.charAt(i)!=str.charAt(len)){
				len = arr[len-1];
			}
		}
	}
	static class ST1{
		int arr[], st[], size;
		ST1(int a[]){
			arr = a.clone();
			size = 10*arr.length;
			st = new int[size];
			build(0,arr.length-1,1);
		}
		void build(int ss, int se, int si){
			if(ss==se){
				st[si] = arr[ss];
				return;
			}
			int mid = (ss+se)/2;
			int val = 2*si;
			build(ss,mid,val); build(mid+1,se,val+1);
			st[si] = Math.min(st[val], st[val+1]);
		}
		int get(int ss, int se, int l, int r, int si){
			if(l>se || r<ss || l>r) return Integer.MAX_VALUE;
			if(l<=ss && r>=se) return st[si];
			int mid = (ss+se)/2;
			int val = 2*si;
			return Math.min(get(ss,mid,l,r,val), get(mid+1,se,l,r,val+1));
		}
	}
	static class ST{
		int arr[],lazy[],n;
		ST(int a){
			n = a;
			arr = new int[10*n];
			lazy = new int[10*n];
		}
		void up(int l,int r,int val){
			update(0,n-1,0,l,r,val);
		}
		void update(int l,int r,int c,int x,int y,int val){
			if(lazy[c]!=0){
				lazy[2*c+1]+=lazy[c];
				lazy[2*c+2]+=lazy[c];
				if(l==r)
					arr[c]+=lazy[c];
				lazy[c] = 0;
			}
			if(l>r||x>y||l>y||x>r)
				return;
			if(x<=l&&y>=r){
				lazy[c]+=val;
				return ;
			}
			int mid = l+r>>1;
			update(l,mid,2*c+1,x,y,val);
			update(mid+1,r,2*c+2,x,y,val);
			arr[c] = Math.max(arr[2*c+1], arr[2*c+2]);
		}
		int an(int ind){
			return ans(0,n-1,0,ind);
		}
		int ans(int l,int r,int c,int ind){
			if(lazy[c]!=0){
				lazy[2*c+1]+=lazy[c];
				lazy[2*c+2]+=lazy[c];
				if(l==r)
					arr[c]+=lazy[c];
				lazy[c] = 0;
			}
			if(l==r)
				return arr[c];
			int mid = l+r>>1;
			if(mid>=ind)
				return ans(l,mid,2*c+1,ind);
			return ans(mid+1,r,2*c+2,ind);
		}
	}
	public static class FenwickTree {
		
	    int[] array; // 1-indexed array, In this array We save cumulative information to perform efficient range queries and updates

	    public FenwickTree(int size) {
	        array = new int[size + 1];
	    }
	    public int rsq(int ind) {
	        assert ind > 0;
	        int sum = 0;
	        while (ind > 0) {
	            sum += array[ind];
	            //Extracting the portion up to the first significant one of the binary representation of 'ind' and decrementing ind by that number
	            ind -= ind & (-ind);
	        }
	        return sum;
	    }
	    public int rsq(int a, int b) {
	        assert b >= a && a > 0 && b > 0;
	        return rsq(b) - rsq(a - 1);
	    }
	    public void update(int ind, int value) {
	        assert ind > 0;
	        while (ind < array.length) {
	            array[ind] += value;
	            //Extracting the portion up to the first significant one of the binary representation of 'ind' and incrementing ind by that number
	            ind += ind & (-ind);
	        }
	    }
	    public int size() {
	        return array.length - 1;
	    }
	}
	public static int[] shuffle(int[] a, Random gen){
		for(int i = 0, n = a.length;i < n;i++)
		{ 
			int ind = gen.nextInt(n-i)+i; 
			int d = a[i]; 
			a[i] = a[ind];
			a[ind] = d; 
		} 
	return a; 
	}

	class Pair implements Comparable<Pair>{
		int u, v;
		Pair(int u, int v){
			this.u = u;
			this.v = v;
		}public int hashCode() {
			return Objects.hash();
		}
		public boolean equals(Object o) {
			Pair other = (Pair) o;
			return ((u == other.u && v == other.v) || (v == other.u && u == other.v));
		}
		public int compareTo(Pair other) {
			//return Integer.compare(val, other.val);
			return Long.compare(u, other.u);// != 0 ? (Long.compare(u, other.u)) : (Long.compare(v,other.v));
		}
	}
	long power(long x, long y, int mod){
		long ans = 1;
		while(y>0){
			if(y%2==0){
				x = (x*x)%mod;
				y/=2;
			}
			else{
				ans = (x*ans)%mod;
				y--;
			}
		}
		return ans;
	}
	void soln() {
		is = System.in;
		out = new PrintWriter(System.out);
		long s = System.currentTimeMillis();
		solve();
		//out.close();
		out.flush();
		//tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Main().soln();
	}

	// To Get Input
	// Some Buffer Methods
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}
	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
									// ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n && !(isSpaceChar(b))) {
			buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;

	private void tr(Object... o) {
		if (!oj)
			System.out.println(Arrays.deepToString(o));
	}
}