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 cf901c {
static ArrayList<Integer>[] adj;
static int[] cnt;
static int[] mx;
static int[] par;
static int tot;
	public static void main(String[] args) throws IOException {
		FastScanner fs=new FastScanner();
		//BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
		//StringTokenizer st=new StringTokenizer(bf.readLine());
		int n=fs.nextInt();
		int m=fs.nextInt();
		PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		adj=new ArrayList[n];
		cnt=new int[n];
		mx=new int[n+1];
		par=new int[n];
		for(int i=0;i<n;i++) {
			adj[i]=new ArrayList<>();
		}
		for(int i=0;i<m;i++) {
			int a1=fs.nextInt()-1;
			int a2=fs.nextInt()-1;
			adj[a1].add(a2);
			adj[a2].add(a1);
		}
		Arrays.fill(mx, Integer.MAX_VALUE-1);
		for(int i=0;i<n;i++) {
			if(cnt[i]==0) {
				dfs(i);
			}
		}
		for(int i=n-1;i>=0;i--) {
			mx[i]=Math.min(mx[i],mx[i+1]);
		}
		long[] pref=new long[n+1];
		for(int i=1;i<=n;i++) {
			pref[i]=pref[i-1]+mx[i]+1-i+1;//converting from zero index to one index
		}
		int q=fs.nextInt();
		while(q-->0) {
			int a1=fs.nextInt();
			int a2=fs.nextInt();
			int l=a1;int r=a2;
			while(l<=r) {
				int md=(l+r)/2;
				if(mx[md]+1>a2) {//0 idx to 1 idx
					r=md-1;
				}
				else {
					l=md+1;
				}
			}
			out.println(pref[r]-pref[a1-1]+(long)(a2-r)*(a2-r+1)/2);
		}
		out.close();
	}
	static void dfs(int x) {
		cnt[x]=++tot;
		for(int i:adj[x]) {
			if(cnt[i]==0) {
				par[i]=x;
				dfs(i);
			}
			else if(cnt[i]>cnt[x]) {
				int u=x;int v=x;
				for(;i!=x;i=par[i]) {
					u=Math.min(u, i);
					v=Math.max(v, i);
				}
				mx[u+1]=Math.min(mx[u+1], v-1);
			}
		}
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}
