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 MAXN = 300100;
vector<int> g[MAXN];
vector<int> gb[MAXN];
bool used[MAXN];
int timer, tin[MAXN], fup[MAXN];
void IS_BRIDGE(int a, int b) {
  gb[a].push_back(b);
  gb[b].push_back(a);
}
int n;
void dfs(int v, int p = -1) {
  used[v] = true;
  tin[v] = fup[v] = timer++;
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to == p) continue;
    if (used[to])
      fup[v] = min(fup[v], tin[to]);
    else {
      dfs(to, v);
      fup[v] = min(fup[v], fup[to]);
      if (fup[to] > tin[v]) IS_BRIDGE(v, to);
    }
  }
}
void find_bridges() {
  timer = 0;
  for (int i = 0; i < n; ++i) used[i] = false;
  for (int i = 0; i < n; ++i)
    if (!used[i]) dfs(i);
}
int clr[300100];
vector<int> c;
void dfs1(int i) {
  clr[i] = 1;
  c.push_back(i);
  for (auto x : g[i]) {
    if (!clr[x] && lower_bound(gb[i].begin(), gb[i].end(), x) ==
                       upper_bound(gb[i].begin(), gb[i].end(), x)) {
      dfs1(x);
    }
  }
}
int far[300100];
long long int kek[300100];
long long int kok[300100];
int lel[300100];
long long int lels[300100];
long long int lol(int a, int b) {
  if (b > n && a < 1) {
    return 0;
  }
  int g = lower_bound(lel + b, lel + n + 1, a) - lel;
  long long int u = a * 1ll * (n + 1 - b) - (lels[g - 1] - lels[b - 1]) -
                    a * 1ll * (n + 1 - g);
  return u;
}
int main() {
  ios_base::sync_with_stdio(false);
  int m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    far[i] = n + 1;
  }
  find_bridges();
  for (int i = 1; i <= n; i++) {
    sort(gb[i].begin(), gb[i].end());
  }
  for (int i = 1; i <= n; i++) {
    if (!clr[i]) {
      dfs1(i);
      if (c.size() > 1) {
        far[*min_element(c.begin(), c.end())] =
            *max_element(c.begin(), c.end());
      }
      c.clear();
    }
  }
  far[n + 1] = n + 1;
  for (int i = n; i >= 1; i--) {
    far[i] = min(far[i], far[i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    ++kek[i];
    --kek[far[i]];
  }
  for (int i = 1; i <= n; i++) {
    kek[i] += kek[i - 1];
    lel[i] = i - kek[i];
    lels[i] = lel[i] + lels[i - 1];
  }
  for (int i = 1; i < n; i++) {
    kek[i + 1] += kek[i];
  }
  for (int i = 1; i <= n; i++) {
    kok[i] = far[i] - i + kok[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    long long int res = kek[b] - kok[a - 1] + lol(a - 1, b + 1);
    cout << res << endl;
  }
  return 0;
}
