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;
using ll = long long;
constexpr int MaxN = 3e5, oo = 1e9;
int n, m;
bool odw[MaxN + 1];
int ojc[MaxN + 1];
int depth[MaxN + 1];
int ilo[MaxN + 2];
ll pref[MaxN + 1];
vector<pair<int, int> > tb;
vector<int> gr[MaxN + 1];
inline void dfs(int v, int oj) {
  ojc[v] = oj;
  odw[v] = true;
  depth[v] = depth[oj] + 1;
  for (int s : gr[v])
    if (s != oj) {
      if (odw[s] == false)
        dfs(s, v);
      else if (depth[s] < depth[v]) {
        int maks = s, mini = s;
        int q = v;
        while (q != s) {
          maks = max(maks, q);
          mini = min(mini, q);
          q = ojc[q];
        }
        tb.push_back({mini, maks});
      }
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 1; i <= n + 1; ++i) ilo[i] = oo;
  for (int i = 0; i < m; ++i) {
    int p, q;
    cin >> p >> q;
    gr[p].push_back(q);
    gr[q].push_back(p);
  }
  for (int i = 1; i <= n; ++i)
    if (not odw[i]) dfs(i, i);
  for (auto& p : tb) ilo[p.first] = min(ilo[p.first], p.second);
  for (int i = n - 1; i >= 1; --i) ilo[i] = min(ilo[i], ilo[i + 1]);
  for (int i = 1; i <= n; ++i) --ilo[i];
  for (int i = 1; i <= n; ++i) pref[i] = pref[i - 1] + ilo[i] - i + 1;
  ;
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    int dl = -1;
    for (int b = n; b >= 1; b >>= 1)
      while (l + dl + b <= r and ilo[l + dl + b] <= r) dl += b;
    ll sum = 0;
    sum += (ll)(r - (l + dl)) * (r - (l + dl) + 1) / 2;
    sum += pref[l + dl] - pref[l - 1];
    cout << sum << '\n';
  }
}
