#include <bits/stdc++.h>
using namespace std;
clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
void _F(const char *name, T arg1) {
  cerr << name << " = " << arg1 << '\n';
}
template <typename T, typename... Args>
void _F(const char *names, T arg1, Args... args) {
  const char *name = strchr(names, ',');
  cerr.write(names, name - names) << " = " << arg1 << '\n';
  _F(name + 2, args...);
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &q) {
  in >> q.first >> q.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &q) {
  out << q.first << " " << q.second;
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first + p2.first, p1.second + p2.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
const long long N = 3e5 + 1;
std::vector<long long> g[N], vis(N);
long long arr[N], par[N], tin[N], t;
long long n, m, a, b;
void dfs(long long v) {
  vis[v] = 1;
  tin[v] = ++t;
  for (auto u : g[v]) {
    if (vis[u] && par[v] != u && tin[u] < tin[v]) {
      long long mx = u, mn = u, cur = v;
      while (cur != u) {
        mx = max(mx, cur);
        mn = min(cur, mn);
        cur = par[cur];
      }
      arr[mn] = mx - 1;
    }
    if (!vis[u]) {
      par[u] = v;
      dfs(u);
    }
  }
}
struct node {
  std::vector<long long> s, sum;
} seg[4 * N];
struct SEGTREE {
  void Add(node l, node r, node &v) {
    long long i = 0, j = 0;
    while (i < (long long)((l.s).size()) && j < (long long)((r.s).size())) {
      if (l.s[i] < r.s[j]) {
        v.s.push_back(l.s[i]);
        i++;
      } else {
        v.s.push_back(r.s[j]);
        j++;
      }
    }
    while (i < (long long)((l.s).size())) {
      v.s.push_back(l.s[i]);
      i++;
    }
    while (j < (long long)((r.s).size())) {
      v.s.push_back(r.s[j]);
      j++;
    }
    for (long long i = 0; i < (long long)((v.s).size()); i++) {
      if (!i)
        v.sum.push_back(v.s[i]);
      else
        v.sum.push_back(v.sum.back() + v.s[i]);
    }
  }
  void Build(long long s, long long e, long long v) {
    if (s == e) {
      seg[v].s.push_back(arr[s]);
      seg[v].sum.push_back(arr[s]);
    } else {
      long long mid = (s + e) >> 1;
      Build(s, mid, 2 * v);
      Build(mid + 1, e, 2 * v + 1);
      Add(seg[2 * v], seg[2 * v + 1], seg[v]);
    }
  }
  long long Query(long long s, long long e, long long v, long long l,
                  long long r) {
    if (r < l || s > r || l > e || s > e) return 0;
    if (l <= s && r >= e) {
      long long j =
          (long long)(lower_bound((seg[v].s).begin(), (seg[v].s).end(), r) -
                      begin(seg[v].s)) -
          1;
      long long x = ((long long)((seg[v].s).size()) - j - 1) * r;
      if (j >= 0)
        return seg[v].sum[j] + x;
      else
        return x;
    }
    long long mid = (s + e) >> 1;
    long long ll = Query(s, mid, 2 * v, l, r);
    long long rr = Query(mid + 1, e, 2 * v + 1, l, r);
    return ll + rr;
  }
} st;
void solve() {
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    arr[i] = n;
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
  }
  for (long long i = n - 1; i > 0; i--) {
    arr[i] = min(arr[i], arr[i + 1]);
  }
  st.Build(1, n, 1);
  long long q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    cout << st.Query(1, n, 1, l, r) - r * (r + 1) / 2 + l * (l - 1) / 2 +
                (r - l + 1)
         << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  start = clock();
  cout << fixed << setprecision(20);
  for (long long i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
void _F(const char *name, T arg1) {
  cerr << name << " = " << arg1 << '\n';
}
template <typename T, typename... Args>
void _F(const char *names, T arg1, Args... args) {
  const char *name = strchr(names, ',');
  cerr.write(names, name - names) << " = " << arg1 << '\n';
  _F(name + 2, args...);
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &q) {
  in >> q.first >> q.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &q) {
  out << q.first << " " << q.second;
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first + p2.first, p1.second + p2.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
const long long N = 3e5 + 1;
std::vector<long long> g[N], vis(N);
long long arr[N], par[N], tin[N], t;
long long n, m, a, b;
void dfs(long long v) {
  vis[v] = 1;
  tin[v] = ++t;
  for (auto u : g[v]) {
    if (vis[u] && par[v] != u && tin[u] < tin[v]) {
      long long mx = u, mn = u, cur = v;
      while (cur != u) {
        mx = max(mx, cur);
        mn = min(cur, mn);
        cur = par[cur];
      }
      arr[mn] = mx - 1;
    }
    if (!vis[u]) {
      par[u] = v;
      dfs(u);
    }
  }
}
struct node {
  std::vector<long long> s, sum;
} seg[4 * N];
struct SEGTREE {
  void Add(node l, node r, node &v) {
    long long i = 0, j = 0;
    while (i < (long long)((l.s).size()) && j < (long long)((r.s).size())) {
      if (l.s[i] < r.s[j]) {
        v.s.push_back(l.s[i]);
        i++;
      } else {
        v.s.push_back(r.s[j]);
        j++;
      }
    }
    while (i < (long long)((l.s).size())) {
      v.s.push_back(l.s[i]);
      i++;
    }
    while (j < (long long)((r.s).size())) {
      v.s.push_back(r.s[j]);
      j++;
    }
    for (long long i = 0; i < (long long)((v.s).size()); i++) {
      if (!i)
        v.sum.push_back(v.s[i]);
      else
        v.sum.push_back(v.sum.back() + v.s[i]);
    }
  }
  void Build(long long s, long long e, long long v) {
    if (s == e) {
      seg[v].s.push_back(arr[s]);
      seg[v].sum.push_back(arr[s]);
    } else {
      long long mid = (s + e) >> 1;
      Build(s, mid, 2 * v);
      Build(mid + 1, e, 2 * v + 1);
      Add(seg[2 * v], seg[2 * v + 1], seg[v]);
    }
  }
  long long Query(long long s, long long e, long long v, long long l,
                  long long r) {
    if (r < l || s > r || l > e || s > e) return 0;
    if (l <= s && r >= e) {
      long long j =
          (long long)(lower_bound((seg[v].s).begin(), (seg[v].s).end(), r) -
                      begin(seg[v].s)) -
          1;
      long long x = ((long long)((seg[v].s).size()) - j - 1) * r;
      if (j >= 0)
        return seg[v].sum[j] + x;
      else
        return x;
    }
    long long mid = (s + e) >> 1;
    long long ll = Query(s, mid, 2 * v, l, r);
    long long rr = Query(mid + 1, e, 2 * v + 1, l, r);
    return ll + rr;
  }
} st;
void solve() {
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    arr[i] = n;
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
  }
  for (long long i = n - 1; i > 0; i--) {
    arr[i] = min(arr[i], arr[i + 1]);
  }
  st.Build(1, n, 1);
  long long q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    cout << st.Query(1, n, 1, l, r) - r * (r + 1) / 2 + l * (l - 1) / 2 +
                (r - l + 1)
         << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  start = clock();
  cout << fixed << setprecision(20);
  for (long long i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

```