#include <bits/stdc++.h>
using namespace std;
vector<array<long long int, 2>> lr_seg;
set<array<long long int, 2>> edge_vis;
bool cmp(array<long long int, 2> a, array<long long int, 2> b) {
  if (a[0] != b[0]) {
    return a[0] < b[0];
  }
  return a[1] > b[1];
}
bool vis[300003];
vector<long long int> v;
vector<long long int> g[300003];
long long int mn_right[300003];
long long int ans[300003];
void dfs(long long int node, long long int parent) {
  vis[node] = 1;
  v.push_back(node);
  for (auto it : g[node]) {
    if (it == parent) {
      continue;
    }
    if (edge_vis.find({min(it, node), max(it, node)}) != edge_vis.end()) {
      continue;
    }
    edge_vis.insert({min(it, node), max(it, node)});
    if (vis[it]) {
      long long int Mn = it;
      long long int Mx = it;
      for (long long int i = v.size() - 1ll; i >= 0 && v[i] != it; i--) {
        Mn = min(Mn, v[i]);
        Mx = max(Mx, v[i]);
      }
      lr_seg.push_back({Mn, Mx});
    }
    if (!vis[it]) {
      dfs(it, node);
    }
  }
  v.pop_back();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 0; i < m; i++) {
    long long int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (long long int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 0);
    }
  }
  sort(lr_seg.begin(), lr_seg.end(), cmp);
  long long int cur = n + 1ll;
  for (long long int i = n; i >= 1; i--) {
    long long int indx = lower_bound(lr_seg.begin(), lr_seg.end(),
                                     array<long long int, 2>{i, 0}) -
                         lr_seg.begin();
    if (indx < lr_seg.size()) {
      cur = min(cur, lr_seg[indx][1]);
    }
    mn_right[i] = cur - 1ll;
  }
  for (long long int i = 1; i <= n; i++) {
    ans[i] = ans[i - 1ll] + (mn_right[i] - i + 1ll);
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    long long int lo = l;
    long long int hi = r;
    long long int indx = n + 1ll;
    while (lo <= hi) {
      long long int mid = (lo + hi) / 2;
      if (mn_right[mid] > r) {
        indx = mid;
        hi = mid - 1ll;
      } else {
        lo = mid + 1ll;
      }
    }
    long long int res = 0;
    if (indx == n + 1ll) {
      res = ans[r] - ans[l - 1ll];
    } else {
      res = ans[indx - 1ll] - ans[l - 1ll];
      res += ((r - indx + 1ll) * (r - indx + 2ll)) / 2;
    }
    cout << res << "\n";
  }
}

### Prompt

In CPP, your task is to solve the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<array<long long int, 2>> lr_seg;
set<array<long long int, 2>> edge_vis;
bool cmp(array<long long int, 2> a, array<long long int, 2> b) {
  if (a[0] != b[0]) {
    return a[0] < b[0];
  }
  return a[1] > b[1];
}
bool vis[300003];
vector<long long int> v;
vector<long long int> g[300003];
long long int mn_right[300003];
long long int ans[300003];
void dfs(long long int node, long long int parent) {
  vis[node] = 1;
  v.push_back(node);
  for (auto it : g[node]) {
    if (it == parent) {
      continue;
    }
    if (edge_vis.find({min(it, node), max(it, node)}) != edge_vis.end()) {
      continue;
    }
    edge_vis.insert({min(it, node), max(it, node)});
    if (vis[it]) {
      long long int Mn = it;
      long long int Mx = it;
      for (long long int i = v.size() - 1ll; i >= 0 && v[i] != it; i--) {
        Mn = min(Mn, v[i]);
        Mx = max(Mx, v[i]);
      }
      lr_seg.push_back({Mn, Mx});
    }
    if (!vis[it]) {
      dfs(it, node);
    }
  }
  v.pop_back();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 0; i < m; i++) {
    long long int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (long long int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 0);
    }
  }
  sort(lr_seg.begin(), lr_seg.end(), cmp);
  long long int cur = n + 1ll;
  for (long long int i = n; i >= 1; i--) {
    long long int indx = lower_bound(lr_seg.begin(), lr_seg.end(),
                                     array<long long int, 2>{i, 0}) -
                         lr_seg.begin();
    if (indx < lr_seg.size()) {
      cur = min(cur, lr_seg[indx][1]);
    }
    mn_right[i] = cur - 1ll;
  }
  for (long long int i = 1; i <= n; i++) {
    ans[i] = ans[i - 1ll] + (mn_right[i] - i + 1ll);
  }
  long long int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    long long int lo = l;
    long long int hi = r;
    long long int indx = n + 1ll;
    while (lo <= hi) {
      long long int mid = (lo + hi) / 2;
      if (mn_right[mid] > r) {
        indx = mid;
        hi = mid - 1ll;
      } else {
        lo = mid + 1ll;
      }
    }
    long long int res = 0;
    if (indx == n + 1ll) {
      res = ans[r] - ans[l - 1ll];
    } else {
      res = ans[indx - 1ll] - ans[l - 1ll];
      res += ((r - indx + 1ll) * (r - indx + 2ll)) / 2;
    }
    cout << res << "\n";
  }
}

```