Problem: Valera loves segments. He has recently come up with one interesting problem.

The Ox axis of coordinates has n segments, the i-th segment starts in position li and ends in position ri (we will mark it as [li, ri]). Your task is to process m queries, each consists of number cnti and a set of cnti coordinates of points located on the Ox axis. The answer to the query is the number of segments, such that each of them contains at least one point from the query. Segment [l, r] contains point q, if l ≤ q ≤ r.

Valera found the solution of this problem too difficult. So he asked you to help him. Help Valera. 

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 3·105) — the number of segments on the axis of coordinates and the number of queries. 

Next n lines contain the descriptions of the segments. The i-th line contains two positive integers li, ri (1 ≤ li ≤ ri ≤ 106) — the borders of the i-th segment.

Next m lines contain the description of the queries, one per line. Each line starts from integer cnti (1 ≤ cnti ≤ 3·105) — the number of points in the i-th query. Then the line contains cnti distinct positive integers p1, p2, ..., pcnti (1 ≤ p1 < p2 < ... < pcnti ≤ 106) — the coordinates of points in the i-th query.

It is guaranteed that the total number of points in all queries doesn't exceed 3·105. 

Output

Print m non-negative integers, where the i-th number is the response to the i-th query.

Examples

Input

3 3
1 3
4 5
6 7
3 1 4 7
2 4 5
1 8


Output

3
1
0

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class tr1 {
	static PrintWriter out;
	static StringBuilder sb;
	static int n, m, k;
	static long mod = 998244353;
	static Boolean[][][] memo;
	static String s;
	static ArrayList<Integer>[] ad, ad1;
	static long inf = Long.MAX_VALUE;
	static int N = 1 << 20 + 1;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int n = sc.nextInt();
		int q = sc.nextInt();
		SegmentTree st = new SegmentTree(new int[N]);
		int[] ans = new int[q];
		int nn = 1000001;
		ad = new ArrayList[nn];
		ad1 = new ArrayList[nn];
		for (int i = 0; i < n; i++) {
			int l = sc.nextInt();
			int r = sc.nextInt();
			if (ad[r] == null)
				ad[r] = new ArrayList<>();
			ad[r].add(l);
		}
		for (int i = 0; i < q; i++) {
			int k = sc.nextInt();
			for (int j = 0; j < k; j++) {
				int x = sc.nextInt();
				if (ad1[x] == null)
					ad1[x] = new ArrayList<>();
				ad1[x].add(i);
			}
		}
		int[] last = new int[q];
		Arrays.fill(last, -1);
//		System.out.println(Arrays.toString(ad1));
	//	System.out.println(Arrays.toString(ad));
		for (int day = 1; day < nn; day++) {
			if (ad1[day] != null)
				for (int v : ad1[day]) {
					if (last[v] != -1) {
						ans[v] += st.query(last[v], last[v]);
					}
					last[v] = day;
				}
			if (ad[day] != null)
				for (int v : ad[day]) {
					st.update_range(v, day, 1);
				}
		}
		for (int i = 0; i < q; i++)
			out.println(ans[i] + st.query(last[i], last[i]));
		out.flush();
	}

	public static class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;

		SegmentTree(int[] in) {
			array = in;
			N = in.length - 1;
			sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N << 1];
			build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;
			sTree[index] += val;
			while (index > 1) {
				index >>= 1;
				sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
			}
		}

		void update_range(int i, int j, int val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, int val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		int query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		int query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			int q1 = query(node << 1, b, mid, i, j);
			int q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return q1 + q2;

		}

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextArrInt(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public long[] nextArrLong(int n) throws IOException {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}