#include <bits/stdc++.h>
using namespace std;
const int MX = 3 * 1e5 + 5;
const int INF = 2 * 1e9 + 5;
int n, m, q;
int mn_end[MX];
long long pref_sum[MX];
vector<vector<int> > g;
vector<pair<int, int> > cycles;
int cycle_detected = 0;
int ref_node = -1;
class UGraphBridgesAP {
  int dfs_cntr;
  vector<int> dfs_num;
  vector<int> dfs_low;
  vector<int> dfs_parent;
  vector<int> articulation_point;
  vector<vector<int> > g;
  int n;
  int dfs_root;
  int root_children;
  set<pair<int, int> > bridges;
  void recurse(int u) {
    dfs_low[u] = dfs_num[u] = dfs_cntr++;
    for (int v : g[u]) {
      if (dfs_num[v] == -1) {
        dfs_parent[v] = u;
        if (u == dfs_root) root_children++;
        recurse(v);
        if (dfs_low[v] >= dfs_num[u]) articulation_point[u] = 1;
        if (dfs_low[v] > dfs_num[u]) bridges.insert({min(u, v), max(u, v)});
        dfs_low[u] = min(dfs_low[u], dfs_low[v]);
      } else if (v != dfs_parent[u])
        dfs_low[u] = min(dfs_low[u], dfs_num[v]);
    }
  }

 public:
  UGraphBridgesAP(vector<vector<int> > _g) {
    g = _g;
    n = g.size();
    dfs_cntr = 0;
    dfs_num.assign(n + 5, -1);
    dfs_low.assign(n + 5, 0);
    dfs_parent.assign(n + 5, 0);
    articulation_point.assign(n + 5, 0);
  }
  void process() {
    for (int i = 0; i < n; i++) {
      if (dfs_num[i] == -1) {
        dfs_root = i;
        root_children = 0;
        recurse(i);
        articulation_point[dfs_root] = (root_children > 1);
      }
    }
  }
  bool is_bridge(int u, int v) {
    if (u > v) swap(u, v);
    return (bridges.find({u, v}) != bridges.end());
  }
  bool is_articulation_point(int u) { return articulation_point[u]; }
};
int minV, maxV;
int vis[MX];
void dfs(int u, int pa, UGraphBridgesAP& ugb) {
  if (vis[u]) {
    cycles.push_back({minV, maxV});
    return;
  }
  vis[u] = 1;
  minV = min(minV, u);
  maxV = max(maxV, u);
  for (int v : g[u]) {
    if (v == pa) continue;
    if (!ugb.is_bridge(u, v)) dfs(v, u, ugb);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  g.resize(MX);
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  scanf("%d", &q);
  UGraphBridgesAP ugb(g);
  ugb.process();
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      minV = MX;
      maxV = 0;
      dfs(i, -1, ugb);
    }
  }
  sort(cycles.begin(), cycles.end());
  int csz = cycles.size();
  for (int i = 1; i <= n; i++) mn_end[i] = n;
  for (int i = 0; i < csz; i++)
    mn_end[cycles[i].first] =
        min(mn_end[cycles[i].first], cycles[i].second - 1);
  for (int i = n - 1; i >= 1; i--) mn_end[i] = min(mn_end[i + 1], mn_end[i]);
  for (int i = 1; i <= n; i++) pref_sum[i] = mn_end[i] - i + 1;
  for (int i = 2; i <= n; i++) pref_sum[i] += pref_sum[i - 1];
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int idx = lower_bound(mn_end + 1, mn_end + n + 1, r) - mn_end;
    idx = max(idx, l);
    idx = min(idx, r);
    long long rem = r - idx + 1;
    long long res = pref_sum[idx - 1] - pref_sum[l - 1] + rem * (rem + 1) / 2;
    printf("%lld\n", res);
  }
  return 0;
}

### Prompt

Your task is to create 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;
const int MX = 3 * 1e5 + 5;
const int INF = 2 * 1e9 + 5;
int n, m, q;
int mn_end[MX];
long long pref_sum[MX];
vector<vector<int> > g;
vector<pair<int, int> > cycles;
int cycle_detected = 0;
int ref_node = -1;
class UGraphBridgesAP {
  int dfs_cntr;
  vector<int> dfs_num;
  vector<int> dfs_low;
  vector<int> dfs_parent;
  vector<int> articulation_point;
  vector<vector<int> > g;
  int n;
  int dfs_root;
  int root_children;
  set<pair<int, int> > bridges;
  void recurse(int u) {
    dfs_low[u] = dfs_num[u] = dfs_cntr++;
    for (int v : g[u]) {
      if (dfs_num[v] == -1) {
        dfs_parent[v] = u;
        if (u == dfs_root) root_children++;
        recurse(v);
        if (dfs_low[v] >= dfs_num[u]) articulation_point[u] = 1;
        if (dfs_low[v] > dfs_num[u]) bridges.insert({min(u, v), max(u, v)});
        dfs_low[u] = min(dfs_low[u], dfs_low[v]);
      } else if (v != dfs_parent[u])
        dfs_low[u] = min(dfs_low[u], dfs_num[v]);
    }
  }

 public:
  UGraphBridgesAP(vector<vector<int> > _g) {
    g = _g;
    n = g.size();
    dfs_cntr = 0;
    dfs_num.assign(n + 5, -1);
    dfs_low.assign(n + 5, 0);
    dfs_parent.assign(n + 5, 0);
    articulation_point.assign(n + 5, 0);
  }
  void process() {
    for (int i = 0; i < n; i++) {
      if (dfs_num[i] == -1) {
        dfs_root = i;
        root_children = 0;
        recurse(i);
        articulation_point[dfs_root] = (root_children > 1);
      }
    }
  }
  bool is_bridge(int u, int v) {
    if (u > v) swap(u, v);
    return (bridges.find({u, v}) != bridges.end());
  }
  bool is_articulation_point(int u) { return articulation_point[u]; }
};
int minV, maxV;
int vis[MX];
void dfs(int u, int pa, UGraphBridgesAP& ugb) {
  if (vis[u]) {
    cycles.push_back({minV, maxV});
    return;
  }
  vis[u] = 1;
  minV = min(minV, u);
  maxV = max(maxV, u);
  for (int v : g[u]) {
    if (v == pa) continue;
    if (!ugb.is_bridge(u, v)) dfs(v, u, ugb);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  g.resize(MX);
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  scanf("%d", &q);
  UGraphBridgesAP ugb(g);
  ugb.process();
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      minV = MX;
      maxV = 0;
      dfs(i, -1, ugb);
    }
  }
  sort(cycles.begin(), cycles.end());
  int csz = cycles.size();
  for (int i = 1; i <= n; i++) mn_end[i] = n;
  for (int i = 0; i < csz; i++)
    mn_end[cycles[i].first] =
        min(mn_end[cycles[i].first], cycles[i].second - 1);
  for (int i = n - 1; i >= 1; i--) mn_end[i] = min(mn_end[i + 1], mn_end[i]);
  for (int i = 1; i <= n; i++) pref_sum[i] = mn_end[i] - i + 1;
  for (int i = 2; i <= n; i++) pref_sum[i] += pref_sum[i - 1];
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int idx = lower_bound(mn_end + 1, mn_end + n + 1, r) - mn_end;
    idx = max(idx, l);
    idx = min(idx, r);
    long long rem = r - idx + 1;
    long long res = pref_sum[idx - 1] - pref_sum[l - 1] + rem * (rem + 1) / 2;
    printf("%lld\n", res);
  }
  return 0;
}

```