You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

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


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>
import java.awt.Point;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;



public class C {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	static ArrayList<Integer> []ages;
	static int[]color;
	static int[]p, last;
	static int[]to;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		
		int n = nextInt();
		int m = nextInt();
		ages = new ArrayList[n+1];
		for (int i = 1; i <= n; i++) {
			ages[i] = new ArrayList<>();
		}
		for (int i = 0; i < m; i++) {
			int v1 = nextInt();
			int v2 = nextInt();
			ages[v1].add(v2);
			ages[v2].add(v1);
		}
		color = new int[n+1];
		p = new int[n+1];
		last = new int[n+1];
		for (int i = 1; i <= n; i++) {
			if (color[i]==0)
				dfs(i);
		}
		int min = n+1;
		to = new int[n+1];
		for (int i = n; i >= 1; i--) {
			if (last[i] > 0)
				min = Math.min(min, last[i]);
			to[i] = min - 1;
		}
		long[]sum = new long[n+1];
		for (int i = 1; i <= n; i++) {
			sum[i] = sum[i-1] + to[i];
		}
		int q = nextInt();

		for (int i = 0; i < q; i++) {
			int L = nextInt();
			int R = nextInt();
			int left = L, right = R+1;
			while (right-left > 1) {
				int mid = (left+right) / 2;
				if (to[mid] > R)
					right = mid;
				else
					left = mid;
			}
			long result = 0;
			if (R >= to[left]) {
				result = sum[left]-sum[L-1];
				result += 1l * R * (R-left);
			}
			else
				result = 1l * R * (R-L+1);
			result -= 1l * (L+R-2) * (R-L+1) / 2;
			pw.println(result);
		}
		pw.close(); 
	}
	private static void dfs(int v) {
		color[v] = 1;
		for (int to : ages[v]) {
			if (to==p[v])
				continue;
			if (color[to]==0) {
				p[to] = v;
				dfs(to);
			}
			else if (color[to]==1) {
				int cur = v;
				int min = v, max = v;
				while (cur != to) {
					cur = p[cur];
					min = Math.min(cur, min);
					max = Math.max(cur, max);
				}
				last[min] = max;
			}
		}
		color[v] = 2;
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}