#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> adj[300009];
vector<pair<int, int> > add[300009];
int sz[300009], comp[300009], S;
int tin[300009], fup[300009], TIM, arr[300009], cnt[300009];
stack<int> st;
long long ans[300009];
void dfs(int nd, int pr) {
  tin[nd] = fup[nd] = ++TIM;
  st.push(nd);
  for (int i = 0; i < int(adj[nd].size()); i++) {
    int to = adj[nd][i];
    if (to == pr) continue;
    if (!tin[to]) {
      dfs(to, nd);
      umin(fup[nd], fup[to]);
    } else
      umin(fup[nd], tin[to]);
  }
  if (fup[nd] == tin[nd]) {
    int aux;
    S++;
    do {
      aux = st.top();
      st.pop();
      comp[aux] = S;
      sz[S]++;
    } while (aux != nd);
  }
}
long long s[300009 << 2];
int lazy[300009 << 2];
void shift(int nd, int x, int y) {
  int& ret = lazy[nd];
  if (!ret) return;
  int mid = (x + y) >> 1;
  s[nd << 1] += (mid - x + 1) * 1LL * ret;
  lazy[nd << 1] += ret;
  s[nd << 1 | 1] += (y - mid) * 1LL * ret;
  lazy[nd << 1 | 1] += ret;
  ret = 0;
}
void inc(int l, int r, int nd, int x, int y) {
  if (l > y or x > r) return;
  if (l <= x and y <= r) {
    lazy[nd]++;
    s[nd] += (y - x + 1);
    return;
  }
  shift(nd, x, y);
  int mid = (x + y) >> 1;
  inc(l, r, nd << 1, x, mid);
  inc(l, r, nd << 1 | 1, mid + 1, y);
  s[nd] = s[nd << 1] + s[nd << 1 | 1];
}
long long tap(int l, int r, int nd, int x, int y) {
  if (l > y or x > r) return 0;
  if (l <= x and y <= r) return s[nd];
  shift(nd, x, y);
  int mid = (x + y) >> 1;
  return tap(l, r, nd << 1, x, mid) + tap(l, r, nd << 1 | 1, mid + 1, y);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!tin[i]) dfs(i, -1);
  int p = 1;
  for (int i = 1; i <= n; i++) {
    if (sz[comp[i]] > 1) {
      cnt[comp[i]]++;
      while (cnt[comp[i]] == sz[comp[i]]) cnt[comp[p]]--, p++;
    }
    arr[i] = p;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    add[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; i++) {
    inc(arr[i], i, 1, 1, n);
    for (__typeof((add[i]).begin()) it = (add[i]).begin(); it != (add[i]).end();
         it++)
      ans[it->second] = tap(it->first, i, 1, 1, n);
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  return 0;
}

### Prompt

Please formulate a CPP solution 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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> adj[300009];
vector<pair<int, int> > add[300009];
int sz[300009], comp[300009], S;
int tin[300009], fup[300009], TIM, arr[300009], cnt[300009];
stack<int> st;
long long ans[300009];
void dfs(int nd, int pr) {
  tin[nd] = fup[nd] = ++TIM;
  st.push(nd);
  for (int i = 0; i < int(adj[nd].size()); i++) {
    int to = adj[nd][i];
    if (to == pr) continue;
    if (!tin[to]) {
      dfs(to, nd);
      umin(fup[nd], fup[to]);
    } else
      umin(fup[nd], tin[to]);
  }
  if (fup[nd] == tin[nd]) {
    int aux;
    S++;
    do {
      aux = st.top();
      st.pop();
      comp[aux] = S;
      sz[S]++;
    } while (aux != nd);
  }
}
long long s[300009 << 2];
int lazy[300009 << 2];
void shift(int nd, int x, int y) {
  int& ret = lazy[nd];
  if (!ret) return;
  int mid = (x + y) >> 1;
  s[nd << 1] += (mid - x + 1) * 1LL * ret;
  lazy[nd << 1] += ret;
  s[nd << 1 | 1] += (y - mid) * 1LL * ret;
  lazy[nd << 1 | 1] += ret;
  ret = 0;
}
void inc(int l, int r, int nd, int x, int y) {
  if (l > y or x > r) return;
  if (l <= x and y <= r) {
    lazy[nd]++;
    s[nd] += (y - x + 1);
    return;
  }
  shift(nd, x, y);
  int mid = (x + y) >> 1;
  inc(l, r, nd << 1, x, mid);
  inc(l, r, nd << 1 | 1, mid + 1, y);
  s[nd] = s[nd << 1] + s[nd << 1 | 1];
}
long long tap(int l, int r, int nd, int x, int y) {
  if (l > y or x > r) return 0;
  if (l <= x and y <= r) return s[nd];
  shift(nd, x, y);
  int mid = (x + y) >> 1;
  return tap(l, r, nd << 1, x, mid) + tap(l, r, nd << 1 | 1, mid + 1, y);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (!tin[i]) dfs(i, -1);
  int p = 1;
  for (int i = 1; i <= n; i++) {
    if (sz[comp[i]] > 1) {
      cnt[comp[i]]++;
      while (cnt[comp[i]] == sz[comp[i]]) cnt[comp[p]]--, p++;
    }
    arr[i] = p;
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    add[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; i++) {
    inc(arr[i], i, 1, 1, n);
    for (__typeof((add[i]).begin()) it = (add[i]).begin(); it != (add[i]).end();
         it++)
      ans[it->second] = tap(it->first, i, 1, 1, n);
  }
  for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  return 0;
}

```