#include <bits/stdc++.h>
using namespace std;
long long seg[500001], match[500001], bit[500001], sum[500001];
long long mp[500001] = {0};
vector<long long> g[500001];
bool see[500001];
long long depth[500001], p[500001];
multiset<long long> ep;
multiset<long long>::iterator it;
void dfs(long long u, long long par, long long d) {
  see[u] = 1;
  p[u] = par;
  depth[u] = d;
  for (long long i = 0; i < g[u].size(); i++) {
    long long v = g[u][i];
    if (!see[v])
      dfs(v, u, d + 1);
    else if (depth[v] < depth[u] && v != par) {
      long long x = u, mn = v, mx = v;
      while (v != x) {
        mn = min(mn, x);
        mx = max(mx, x);
        x = p[x];
      }
      mp[mn] = mx;
      ep.insert(mx);
    }
  }
  return;
}
void update(long long i, long long val) {
  while (i < 500001) {
    bit[i] += val;
    i += i & (-i);
  }
}
long long get(long long i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & (-i);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, e;
  cin >> n >> e;
  for (long long i = 0; i < e; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(see, 0, sizeof see);
  for (long long i = 1; i <= n; i++) {
    if (!see[i]) {
      dfs(i, 0, 0);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (ep.size()) {
      it = ep.begin();
      seg[i] = *it - 1;
    } else
      seg[i] = n;
    if (mp[i] != 0) ep.erase(ep.find(mp[i]));
  }
  memset(sum, 0, sizeof sum);
  memset(bit, 0, sizeof bit);
  for (long long i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + seg[i] - i + 1;
  }
  int q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    long long ans = 0;
    if (seg[r] > r) {
      long long* id = lower_bound(seg + l, seg + r, r + 1);
      long long temp = (id - seg);
      ans += (r - temp + 1) * (r - temp + 2) / 2;
      ans += (sum[temp - 1] - sum[l - 1]);
    } else
      ans = sum[r] - sum[l - 1];
    cout << ans << endl;
  }
  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;
long long seg[500001], match[500001], bit[500001], sum[500001];
long long mp[500001] = {0};
vector<long long> g[500001];
bool see[500001];
long long depth[500001], p[500001];
multiset<long long> ep;
multiset<long long>::iterator it;
void dfs(long long u, long long par, long long d) {
  see[u] = 1;
  p[u] = par;
  depth[u] = d;
  for (long long i = 0; i < g[u].size(); i++) {
    long long v = g[u][i];
    if (!see[v])
      dfs(v, u, d + 1);
    else if (depth[v] < depth[u] && v != par) {
      long long x = u, mn = v, mx = v;
      while (v != x) {
        mn = min(mn, x);
        mx = max(mx, x);
        x = p[x];
      }
      mp[mn] = mx;
      ep.insert(mx);
    }
  }
  return;
}
void update(long long i, long long val) {
  while (i < 500001) {
    bit[i] += val;
    i += i & (-i);
  }
}
long long get(long long i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[i];
    i -= i & (-i);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, e;
  cin >> n >> e;
  for (long long i = 0; i < e; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(see, 0, sizeof see);
  for (long long i = 1; i <= n; i++) {
    if (!see[i]) {
      dfs(i, 0, 0);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (ep.size()) {
      it = ep.begin();
      seg[i] = *it - 1;
    } else
      seg[i] = n;
    if (mp[i] != 0) ep.erase(ep.find(mp[i]));
  }
  memset(sum, 0, sizeof sum);
  memset(bit, 0, sizeof bit);
  for (long long i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + seg[i] - i + 1;
  }
  int q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    long long ans = 0;
    if (seg[r] > r) {
      long long* id = lower_bound(seg + l, seg + r, r + 1);
      long long temp = (id - seg);
      ans += (r - temp + 1) * (r - temp + 2) / 2;
      ans += (sum[temp - 1] - sum[l - 1]);
    } else
      ans = sum[r] - sum[l - 1];
    cout << ans << endl;
  }
  return 0;
}

```