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>
#include <bits/stdc++.h>
using namespace std;
const int M = 3e5 + 5, LG = 26, SM = 600, inf = 1e9 + 1;
int mod = 95542721;
int n, m;
int l[M];
int r[M];
vector<int> g[M];
int par[M];
int h[M];
void cyc(int a, int b) {
  int mi = b, mx = b;
  while (a != b) {
    mi = min(mi, a);
    mx = max(mx, a);
    a = par[a];
    if (a == -2) exit(0);
  }
  l[mi] = min(l[mi], mx);
  r[mx] = max(r[mx], mi);
}
void dfs(int x) {
  for (int y : g[x]) {
    if (y == par[x])
      continue;
    else if (par[y] != -1 && h[y] < h[x])
      cyc(x, y);
    else if (par[y] == -1) {
      h[y] = h[x] + 1;
      par[y] = x;
      dfs(y);
    }
  }
}
int cul, cur;
long long int res = 0;
inline void adl() {
  cul--;
  res += min(cur, l[cul]) - cul;
}
inline void del() {
  res -= min(cur, l[cul]) - cul;
  cul++;
}
inline void adr() {
  res += cur - max(cul - 1, r[cur]);
  cur++;
}
inline void der() {
  cur--;
  res -= cur - max(cul - 1, r[cur]);
}
int ql[M], qr[M], qj[M];
long long int qa[M];
bool mo(int i, int j) {
  if (ql[i] / SM == ql[j] / SM) return (qr[i] < qr[j]) ^ ((ql[i] / SM) & 1);
  return ql[i] < ql[j];
}
int main() {
  scanf("%d%d", &n, &m);
  fill(l, l + n, inf);
  fill(r, r + n, -inf);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  fill(par, par + n, -1);
  for (int i = 0; i < n; i++)
    if (par[i] == -1) {
      par[i] = -2;
      dfs(i);
    }
  for (int i = n - 2; i >= 0; i--) {
    l[i] = min(l[i], l[i + 1]);
  }
  for (int i = 1; i < n; i++) r[i] = max(r[i], r[i - 1]);
  int q;
  cin >> q;
  cul = 0;
  cur = 1;
  res = 1;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &ql[i], &qr[i]);
    ql[i]--;
  }
  iota(qj, qj + q, 0);
  sort(qj, qj + q, mo);
  for (int ii = 0; ii < q; ii++) {
    int i = qj[ii];
    int l = ql[i];
    int r = qr[i];
    while (cul > l) adl();
    while (cur < r) adr();
    while (cul < l) del();
    while (cur > r) der();
    qa[i] = res;
  }
  for (int i = 0; i < q; i++) printf("%lld\n", qa[i]);
}
