#include <bits/stdc++.h>
using namespace std;
int n, m, curr = -1;
vector<int> adj[300001];
vector<int> st;
set<int> cycle[300001];
int lt[300001];
bool visited[300001], added[300001];
int bs(int l, int r) {
  int x = r, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (lt[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return l - 1;
}
void dfs(int node = 1, int pa = -1) {
  if (visited[node] && added[node]) {
    curr++;
    while (st.size() && st.back() != node) {
      cycle[curr].insert(st.back());
      added[st.back()] = true;
      st.pop_back();
    }
    st.pop_back();
    cycle[curr].insert(node);
    added[node] = true;
    return;
  }
  visited[node] = true;
  added[node] = true;
  for (auto it : adj[node]) {
    if (it == pa || (visited[it] && !added[it])) continue;
    if (st.size() == 0 || st.back() != node) st.push_back(node);
    dfs(it, node);
  }
  if (st.back() == node) st.pop_back();
  added[node] = false;
}
int main() {
  int a, b;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i)
    if (!visited[i]) dfs(i);
  for (int i = 1; i <= n; ++i) lt[i] = n;
  for (int i = 0; i <= curr; ++i) {
    set<int>::iterator it = cycle[i].begin();
    set<int>::iterator it2 = cycle[i].end();
    it2--;
    lt[*it] = (*it2) - 1;
    it++;
    while (it != cycle[i].end()) {
      lt[*it] = n;
      it++;
    }
  }
  lt[n] = n;
  for (int i = n - 1; i > 0; --i) lt[i] = min(lt[i], lt[i + 1]);
  vector<long long> presum(n + 1, 0);
  for (int i = 1; i <= n; ++i) {
    presum[i] = presum[i - 1] + (long long)lt[i] - (long long)i + 1LL;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int rig = bs(l, r);
    long long ans = presum[rig] - presum[l - 1];
    long long temp = ((long long)r * ((long long)r + 1LL)) / 2LL;
    temp -= (((long long)rig + 1LL) * ((long long)rig)) / 2LL;
    ans += (long long)r * ((long long)r - (long long)rig) - temp +
           ((long long)r - (long long)rig);
    printf("%lld\n", ans);
  }
}

### 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;
int n, m, curr = -1;
vector<int> adj[300001];
vector<int> st;
set<int> cycle[300001];
int lt[300001];
bool visited[300001], added[300001];
int bs(int l, int r) {
  int x = r, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (lt[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return l - 1;
}
void dfs(int node = 1, int pa = -1) {
  if (visited[node] && added[node]) {
    curr++;
    while (st.size() && st.back() != node) {
      cycle[curr].insert(st.back());
      added[st.back()] = true;
      st.pop_back();
    }
    st.pop_back();
    cycle[curr].insert(node);
    added[node] = true;
    return;
  }
  visited[node] = true;
  added[node] = true;
  for (auto it : adj[node]) {
    if (it == pa || (visited[it] && !added[it])) continue;
    if (st.size() == 0 || st.back() != node) st.push_back(node);
    dfs(it, node);
  }
  if (st.back() == node) st.pop_back();
  added[node] = false;
}
int main() {
  int a, b;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i)
    if (!visited[i]) dfs(i);
  for (int i = 1; i <= n; ++i) lt[i] = n;
  for (int i = 0; i <= curr; ++i) {
    set<int>::iterator it = cycle[i].begin();
    set<int>::iterator it2 = cycle[i].end();
    it2--;
    lt[*it] = (*it2) - 1;
    it++;
    while (it != cycle[i].end()) {
      lt[*it] = n;
      it++;
    }
  }
  lt[n] = n;
  for (int i = n - 1; i > 0; --i) lt[i] = min(lt[i], lt[i + 1]);
  vector<long long> presum(n + 1, 0);
  for (int i = 1; i <= n; ++i) {
    presum[i] = presum[i - 1] + (long long)lt[i] - (long long)i + 1LL;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    int rig = bs(l, r);
    long long ans = presum[rig] - presum[l - 1];
    long long temp = ((long long)r * ((long long)r + 1LL)) / 2LL;
    temp -= (((long long)rig + 1LL) * ((long long)rig)) / 2LL;
    ans += (long long)r * ((long long)r - (long long)rig) - temp +
           ((long long)r - (long long)rig);
    printf("%lld\n", ans);
  }
}

```