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.io.*;
import java.util.*;

public class Main {

	
	
	private void solve() throws Exception {

		this.n = nextInt();
		this.m = nextInt();
		
		adj = new LinkedList[n];
		for(int i = 0; i < n; i++) adj[i] = new LinkedList<Integer>();
		
		for(int i = 0; i < m; i++){
			int u = nextInt()-1;
			int v = nextInt()-1;
			adj[u].addLast(v);
			adj[v].addLast(u);
		}

		f = new long[n];
		Arrays.fill(f,-1);
		dfs();
		for(int i = 1; i < n; i++){
			f[i] = Math.max(f[i], f[i-1]);
		}
		
		for(int i = 0; i < n; i++) f[i]++;
		
		
		h = new long[n];
		h[0] = f[0];
		for(int i = 1;i < n; i++) h[i] = h[i-1]+f[i];

		g = new int[n];
		int idx = 0;
		for(int i = 0; i < n; i++) {
			while(idx < n && f[idx] < i) idx++;
			g[i] = idx-1;
		}
		
		int q = nextInt();
		for(int i = 0; i < q; i++){
			int l = nextInt()-1;
			int r = nextInt()-1;
			long ll = (long)l;
			long rr = (long)r;
			long A = (ll+rr+2)*(rr-ll+1)/2L;
			long B = ll*(long)Math.max(0,  Math.min(rr,g[l])-ll+1L);
			long C = Math.max(0L,h(r) - h(g[l]));
			out.println(A-B-C);
		}
	}
	
	
	int n;
	int m;
	
	long h(int i) {
		if(i < 0) return 0;
		else return h[i];
	}
	
	long[] h;	
	int[] g;
	
	LinkedList<Integer>[] adj;
	
	int[] p;
	long[] f;
	
	int[] color;
	
	void dfs(){
		color = new int[n];
		p = new int[n];
		Arrays.fill(p, -1);		
		for(int i = 0; i < n; i++) if(color[i] == 0) visit(i);
	}
	
	void visit(int u){
		color[u] = 1;
		for(Integer v : adj[u]){
			if(color[v] == 0){
				p[v] = u;
				visit(v);
			}else if(color[v] == 1 && p[u] != v){
				int min = u;
				int max = u;
				int w = u;
				while(w != v){
					w = p[w];
					min = Math.min(min, w);
					max = Math.max(max, w);
				}
				f[max] = min;
			}
		}
		color[u] = 2;
	}
	
	
	public static void main(String[] args) {
		new Main().run();
	}

	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer tokenizer;

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			tokenizer = null;
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	private long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	private float nextFloat() throws IOException {
		return Float.parseFloat(nextToken());
	}

	private String nextLine() throws IOException {
		return new String(in.readLine());
	}

	private String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(in.readLine());
		}
		return tokenizer.nextToken();
	}
}
